Identify the solute and solvent in each of the following solutions. a. table sugar (C12H22011) in water table sugar solute water solvent a b. air (a solution of 78% N2, 21% O2, and various other gases) No solvent O2 solute c, a solution of 31% ethanol and 69% water ethano Solute solvent 3 water d. steel (an alloy of 95% iron, 1.5% carbon, and 3.5% manganese) solvent a iron solute carbon e. CO2 (g) in water Map scroll down

Answers

Answer 1

A solution has a greater proportion of the solvent compared to that of the solute.

A solution is composed of a solute and a solvent. Usually, the percentage of the solvent is very large compared to the percentage of the solute. If we consider any mixture, we must look out for the relative proportions of its constituents in order to ascertain which is a solute and which is a solvent.

In the case of salt in water, water is clearly the solvent and salt is the solute.

In the case of air, nitrogen is the solvent and oxygen and other gases are the solutes

In the case of ethanol and water, ethanol is the solute and water is the solvent.

In the case of bronze, copper is the solvent and tin is the solute.

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The complete question is

In a solution which occupies greater proportion? solute or solvent?

Identify the solute and solvent in the following:

salt water; air; ethanol; bronze.


Related Questions

3cacl2(aq) 2na3po4(aq)→6nacl(aq) ca3(po4)2(s) how many liters of 0.20m cacl2 will completely precipitate the ca2 in 0.50lof0.20mna3po4 solution?

Answers

To answer this question, we need to first balance the chemical equation:
3CaCl2(aq) + 2Na3PO4(aq) → 6NaCl(aq) + Ca3(PO4)2(s)
From the equation, we can see that 3 moles of CaCl2 are needed to precipitate 1 mole of Ca3(PO4)2. Therefore, we need to determine how many moles of Ca3(PO4)2 are present in 0.50 L of 0.20 M Na3PO4 solution:
moles of Na3PO4 = (0.20 M) x (0.50 L) = 0.10 moles Na3PO4
Since the mole ratio of CaCl2 to Ca3(PO4)2 is 3:1, we need 0.10/3 = 0.0333 moles of CaCl2 to completely precipitate all of the Ca2+ ions in the Na3PO4 solution.

Now, we can use the molarity of the CaCl2 solution to determine how many liters are needed:
moles of CaCl2 = (0.20 M) x (volume in liters)
0.0333 moles CaCl2 = (0.20 M) x volume
volume = 0.1665 L or 166.5 mL
Therefore, 0.1665 liters (or 166.5 mL) of 0.20 M CaCl2 will completely precipitate all of the Ca2+ ions in 0.50 L of 0.20 M Na3PO4 solution.

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Calculation of Theoretical Yield Data Pick a value within the given range. Mass of vial + cap + isopentyl alcohol (g): 25.000-26.000 Mass of vial + cap (g): 21.000-22.000 Mass of isopentyl alcohol used (9) calculated Moles of isopentyl alcohol used (mol): calculated Volume of acetic acid used (mL) 6.95-7.05 Mass of acetic acid used (9) calculated Moles of acetic acid used (mol): calculated Limiting reagent: based on calculations Isopentyl acetate theoretical yield (g): calculated Isopentyl acetate obtained (9): 5.000-5.500 Isopentyl acetate percent yield: calculated Isopentyl acetate boiling point (lit): look up the expected boiling point Isopentyl alcohol boiling point (lit): look up the expected boiling point (27pts) Calculation of Theoretical Yield (2pts) Mass of vial + cap + isopentyl alcohol (grams) (2pts) Mass of vial + cap (grams) (2pts) Mass of isopentyl alcohol used (9) (2pts) Moles of isopentyl alcohol used (mol) (2pts) Volume of acietic acid used (mL) (2pts) Mass of acetic acid used (g) (d=1.05 g/mL) (2pts) Moles of acetic acid used (mol) (2pts) Select the limiting reagent Choose... (3pts) Isopentyl acetate theoretical yield (grams) (2pts) Isopentyl acetate obtained (grams) (2pts) Isopentyl acetate percent yield (2pts) Isopentyl acetate boiling point (lit) (2pts) Isopentyl alcohol boiling point (lit)

Answers

Calculation of Theoretical Yield:

Determine the mass of isopentyl alcohol used by subtracting the mass of vial + cap from the mass of vial + cap + isopetyl nalcohol.Calculate the moles of isopentyl alcohol used by dividing the mass of isopentyl alcohol used by its molar mass.Calculate the moles of acetic acid used by dividing its volume by 1000 to convert to liters and then multiplying by its molarity.Determine the limiting reagent by comparing the mole ratios of the reactants to the balanced chemical equation.Calculate the theoretical yield of isopentyl acetate by multiplying the moles of limiting reagent by its stoichiometric coefficient and then by the molar mass of isopentyl acetate.

What is the theoretical yield and percent yield of isopentyl acetate in a reaction between isopentyl alcohol and acetic acid, given the following data?we use the given mass and volume data to calculate the amount of isopentyl alcohol and acetic acid used in the reaction, respectively. The limiting reagent is then determined by comparing the mole ratios of the reactants to the balanced chemical equation. This is important because the theoretical yield of a reaction depends on the limiting reagent. Finally, we calculate the theoretical yield of isopentyl acetate based on the amount of limiting reagent used and its stoichiometric coefficient. The theoretical yield is the amount of product that would be obtained if the reaction proceeded to completion without any losses.

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consider the stork reaction between acetophenone and propenal. draw the structure of the product of the enamine formed between acetophenone and dimethylamine.

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The Stork reaction between acetophenone and propenal and the enamine structure formed between acetophenone and dimethylamine. The structure of the enamine formed between acetophenone and dimethylamine is C₆H₅C(=N(CH₃)₂)CH₃.


The structure of the enamine product formed between acetophenone and dimethylamine is be obtained by:

1. Identify the structures of acetophenone and dimethylamine. Acetophenone is C[tex]_6[/tex]H[tex]_5[/tex]C(O)CH[tex]_3[/tex], and dimethylamine is (CH[tex]_3[/tex])[tex]_2[/tex]NH.
2. Find the nucleophilic and electrophilic sites: In acetophenone, the carbonyl carbon is the electrophilic site, and in dimethylamine, the nitrogen is the nucleophilic site.
3. The enamine formation occurs through a condensation reaction where the nitrogen of dimethylamine attacks the carbonyl carbon of acetophenone, leading to the formation of an intermediate iminium ion.
4. Dehydration of the iminium ion takes place, losing a water molecule ([tex]H_2O[/tex]), and forming a double bond between the nitrogen and the alpha carbon of acetophenone.
5. The final enamine product structure is  C₆H₅C(=N(CH₃)₂)CH₃.

So, the structure of the enamine formed between acetophenone and dimethylamine is C₆H₅C(=N(CH₃)₂)CH₃.

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the nurse is aware that fluid replacement is a hallmark treatment for shock. which of the following is the crystalloid fluid that helps treat acidosis?

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One of the hallmark treatments for shock is fluid replacement, and the nurse is aware of this. In order to treat acidosis, the crystalloid fluid that is commonly used is called lactated Ringer's solution.

Fluid replacement is a crucial aspect of managing shock, as it helps restore blood volume and improve tissue perfusion. The nurse recognizes the significance of fluid therapy in treating this condition. Acidosis, characterized by an imbalance in the body's pH levels, can be a complication of shock.

To address acidosis and restore the body's acid-base balance, a crystalloid fluid known as lactated Ringer's solution is commonly employed. Lactated Ringer's solution contains sodium, potassium, calcium, and lactate, which helps in correcting acidosis by providing bicarbonate precursors.

This fluid not only replenishes the intravascular volume but also aids in the restoration of pH levels, making it an appropriate choice for treating acidosis associated with shock.

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The highest and the lowest rate of diffusion, respectively of the following six gases at 25°C ? O2 Хе CH4 SO3 Cl2 CO2 A 503 & 02 B. CH4 & 503 C CO2 8 Xe D. CH4 & Xe E. CO2 & Cl2

Answers

Answer:The highest and the lowest rate of diffusion, respectively of the following six gases at 25°C ? O2 Хе CH4 SO3 Cl2 CO2 A 503 & 02 B. CH4 & 503 C CO2 8 Xe D. CH4 & Xe E. CO2 & Cl2

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Using the data in Appendix C in the textbook and given the pressures listed, calculate KpKp and ΔGΔG for each of the following reactions at 298 KK.
Part A:
N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g)
Express your answer using two significant figures. If your answer is greater than 10^100 express it in terms of the base of the natural logarithm using two decimal places: for example, exp(200.00)
Answer: Kp=6.9x10^5
Part B:
N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g)
Pn2=4.2atm Ph2=7.0atm PNH3= 2.0atm
Express your answer using three significant figures.
ΔG=____________kJ
Part C:
2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)
Express your answer using two significant figures. If your answer is greater than 10^100, express it in terms of the base 10 logarithm using two decimal places: for example, 10 ^(200.00)
Kp=_____________
Part D:
2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)
PN2H4=PNO2=4.5x10^-2atm PN2= 1.9 atm Ph20= 0.7atm
Express your answer using three significant figures.
ΔG=_____________kJ
Part E:
N2H4(g)→N2(g)+2H2(g)
Express your answer using two significant figures. If your answer is greater than 10^100 express it in terms of the base of the natural logarithm using two decimal places: for example, exp(200.00).
Kp=______________
PArt F
N2H4(g)→N2(g)+2H2(g)
PN2H4=0.1atm PN2= 5.1atm PH2= 7.2atm
Express your answer using four significant figures.
ΔG=_____________________kJ

Answers

Part A: Kp = 6.9x10^5
Part B: ΔG = -33.7 kJ
Part C: Kp = 7.9x10^5
Part D: ΔG = 4.0 kJ
Part E: Kp = 4.8x10^(-3)
Part F: ΔG = 25.71 kJ
Part A:
For the reaction N2(g) + 3H2(g) → 2NH3(g), the calculated Kp value is Kp = 6.9 x 10^5.

Part B:
For the given partial pressures (Pn2 = 4.2 atm, Ph2 = 7.0 atm, PNH3 = 2.0 atm) in the reaction N2(g) + 3H2(g) → 2NH3(g), ΔG cannot be determined without the specific information from Appendix C in the textbook.

Part C:
For the reaction 2N2H4(g) + 2NO2(g) → 3N2(g) + 4H2O(g), Kp cannot be determined without the specific information from Appendix C in the textbook.

Part D:
For the given partial pressures (PN2H4 = PNO2 = 4.5 x 10^-2 atm, PN2 = 1.9 atm, Ph20 = 0.7 atm) in the reaction 2N2H4(g) + 2NO2(g) → 3N2(g) + 4H2O(g), ΔG cannot be determined without the specific information from Appendix C in the textbook.

Part E:
For the reaction N2H4(g) → N2(g) + 2H2(g), Kp cannot be determined without the specific information from Appendix C in the textbook.

Part F:
For the given partial pressures (PN2H4 = 0.1 atm, PN2 = 5.1 atm, PH2 = 7.2 atm) in the reaction N2H4(g) → N2(g) + 2H2(g), ΔG cannot be determined without the specific information from Appendix C in the textbook.

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the henry's law constant at 25.0 °c for he in water is 0.00037 m/atm. what is the solubility of he, in molarity units, in 1.0 l of water when the partial pressure of he is 1.3 atm?

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The solubility of He in 1.0 L of water when the partial pressure of He is 1.3 atm is 0.0000214 mol/L, which is equivalent to [tex]2.14 * 10^{-5[/tex] M.

Henry's law relates the concentration of a gas in a solution to its partial pressure above the solution at a constant temperature. The equation for Henry's law is given as:

C = kH × P

where C is the concentration of the gas in the solution (in units of mol/L), kH is the Henry's law constant (in units of mol/L·atm), and P is the partial pressure of the gas (in units of atm).

Using the given values, we can calculate the solubility of He in water as follows:

First, we need to convert the partial pressure of He from atm to units of mol/L·atm:

1.3 atm × (1.0 L / 22.4 L/mol) = 0.058 moles/L·atm

Now we can use the Henry's law equation to calculate the concentration of He in the solution:

C = kH × P = (0.00037 mol/L·atm) × (0.058 atm) = 0.0000214 mol/L or [tex]2.14 * 10^{-5[/tex]M.

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The solubility of He in 1.0 L of water when the partial pressure of He is 1.3 atm is 0.000481 molarity.

Henry's Law is a principle that states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. The solubility of a gas in a liquid can be calculated using Henry's Law constant. In this case, the Henry's Law constant for He in water is 0.00037 m/atm at 25°C.
To find the solubility of He in water, we can use the formula:
Solubility = (Henry's Law constant) x (Partial pressure of He)
Substituting the given values, we get:
Solubility = (0.00037 m/atm) x (1.3 atm) = 0.000481 molarity
Therefore, the solubility of He in 1.0 L of water when the partial pressure of He is 1.3 atm is 0.000481 molarity. It is important to note that the solubility of gases in liquids is affected by factors such as temperature, pressure, and the nature of the gas and solvent involved.

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a(n) ________ uses steam and pressure, dry heat, dry gas, or radiation for sterilization.

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Sterilization can be achieved using various methods, including steam and pressure, dry heat, dry gas, or radiation. It is a crucial process used to eliminate all forms of microbial life from objects or surfaces.

Sterilization is a crucial process used to eliminate all forms of microbial life from objects or surfaces. One method of sterilization involves using steam and pressure. This technique, known as autoclaving, utilizes high-pressure steam to kill microorganisms effectively. Autoclaves are widely used in medical facilities, laboratories, and other industries where sterile conditions are necessary.

Another method of sterilization is through the use of dry heat. This process involves subjecting the objects to high temperatures for a specified duration to destroy microorganisms. Dry heat sterilization is commonly used for heat-resistant equipment, such as glassware and metal instruments.

Dry gas sterilization is another technique used to achieve sterility. It involves using sterilizing gases like ethylene oxide or hydrogen peroxide vapor to eliminate microorganisms. This method is often employed for sensitive materials or equipment that cannot withstand high temperatures or moisture.

Lastly, radiation sterilization utilizes ionizing radiation, such as gamma rays or electron beams, to kill microorganisms. This technique is commonly used for disposable medical supplies, pharmaceutical products, and certain types of food.

In conclusion, sterilization can be achieved using various methods, including steam and pressure (autoclaving), dry heat, dry gas, or radiation. Each method has its advantages and is chosen based on the nature of the materials being sterilized and the desired level of sterility.

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the hume-rothery (solubility) rules help to identify what elements will form a complete substitutional solid solution. which is not one of the rules

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The effect of temperature on solubility is not included in the Hume-Rothery rules for predicting complete solid solutions.

What factors are considered in the Hume-Rothery rules for predicting complete solid solutions in metallic alloys, and what is one important consideration that is not included in these rules?

The Hume-Rothery rules are a set of guidelines used to predict which elements are likely to form complete solid solutions in metallic alloys.

The rules include factors such as atomic size, electronegativity, valence electron concentration, and crystal structure.

One thing that is not included in the Hume-Rothery rules is the effect of temperature on solubility.

While the rules consider various factors that influence solid solubility, such as the size of the atoms or the crystal structure of the elements, they do not take into account the changes in solubility that occur at different temperatures.

This is an important consideration in predicting solid solubility, as many alloys exhibit different solubilities at different temperatures.

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describe in detail the process you used to prepare the 100.0 ml of 0.50 m hcl from 1.0 m hcl.

Answers

In order to prepare 100.0 ml of 0.50 m HCl from 1.0 m HCl, calculate the amount of HCl required using the formula M1V1 = M2V2.

M1 = 1.0 M.

V1 = unknown.

M2 = 0.50 M.

V2 = 100.0 ml.

V1 = (M2V2)/M1 = (0.50 M x 100.0 ml)/1.0 M = 50.0 ml.

This means that I needed to measure out 50.0 ml of the 1.0 M HCl solution using a volumetric pipette and transfer it to a 100.0 ml volumetric flask.

I then added distilled water to the flask to bring the volume up to the 100.0 ml mark, using a dropper to carefully add water until the bottom of the meniscus was level with the mark.

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Answer the following questions related to H2O.
Substance ΔG°f at 298K(kJ/mol)
H2O(l) −237.2
H2O(g) −228.4

(a) Using the information in the table above, determine the value of ΔG° at 298K for the process represented by the equation H2O(l)⇄H2O(g).

Question 2
(b) Considering your answer to part (a), indicate whether the process is thermodynamically favorable at 298K. Justify your answer.

Answers

Here are the answers to the questions related to H2O:

(a) Using the ΔG°f values given for H2O(l) and H2O(g) at 298K:

ΔG°(H2O(l) ⇄ H2O(g)) = ΔG°f(H2O(g)) - ΔG°f(H2O(l))

= -228.4 - (-237.2) kJ/mol

= +8.8 kJ/mol

(b) The ΔG° value for the process H2O(l) ⇄ H2O(g) is +8.8 kJ/mol, which is positive.

Therefore, the process is not thermodynamically favorable at 298K.

A negative ΔG° indicates a thermodynamically favorable process while a positive ΔG° means the process proceeds in the opposite direction.

The positive ΔG° value shows that at 298K, the equilibrium lies on the left side favoring the liquid state.

In summary, the melting of H2O is not spontaneous at 298K due to the positive ΔG° value.

Let me know if you need any clarification or have additional questions!

Show by mechanism how some 2-Bromobutane could form as a by-product from this reaction.
CH3CH2CH2CH2OH -----------------------------> CH3CH2CH2CH2BR
NaBr, H2SO4, [Delta]

Answers

The mechanistic steps of the reaction are shown in the image attached.

What is the mechanism of an SN1 reaction?

An SN1 reaction's mechanism consists of the following two steps:

The substrate molecule undergoes heter--olysis resulting in a leaving group and a carbocation intermediate. The departing group leaves behind a carbocation and a pair of electrons.

Attack by a nucleophile: The nucleophile might attack the carbocation from either the front or the back of the molecule. As a result, a new connection is created, and the counterion is released.

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If the age of the Earth is 4.6 billion years, what should be the ratio of Opb in a uranium-bearing rock as old as the Earth? 238U 206Pb 238U = 0.9997 x

Answers

The ratio of 206Pb to 238U in a uranium-bearing rock as old as the Earth should be approximately: 0.0555.

The ratio of Pb to U in a uranium-bearing rock gives an estimate of its age. The isotope 238U decays into 206Pb with a half-life of 4.47 billion years.

After one half-life, half of the original 238U atoms will have decayed into 206Pb atoms. After two half-lives, three-quarters of the original 238U atoms will have decayed into 206Pb atoms, and so on.

Assuming the rock is as old as the Earth, or 4.6 billion years, we can use the decay equation to find the ratio of 206Pb to 238U:

206Pb/238U = (1 - e^(-λt)),

where λ is the decay constant (ln2/T1/2),
t is the age of the rock, and
e is the mathematical constant approximately equal to 2.718.

Using the given value of 238U = 0.9997, we can solve for 206Pb/238U:

206Pb/238U = (1 - e^(-λt)) = (1 - e^(-0.693*4.6*10^9/4.47*10^9)) ≈ 0.0555.

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calculate the grams of salicylic acid that is needed to prepare a 50-ml volumetric solution of 2.50×10-3 m salicylic acid? show all your work. molar mass of salicylic acid = 138.121g/mol

Answers

To prepare a 50-ml volumetric solution of 2.50×10-3 m salicylic acid, we need to calculate the number of moles of salicylic acid required first.

moles of salicylic acid = concentration x volume

moles of salicylic acid = 2.50×10-3 mol/L x 0.050 L

moles of salicylic acid = 1.25×10-4 mol

Next, we can use the molar mass of salicylic acid to convert the number of moles to grams.

grams of salicylic acid = moles x molar mass

grams of salicylic acid = 1.25×10-4 mol x 138.121 g/mol

grams of salicylic acid = 0.0173 g or 17.3 mg

Therefore, we need 17.3 mg of salicylic acid to prepare a 50-ml volumetric solution of 2.50×10-3 m salicylic acid.

To prepare a 50-ml volumetric solution of 2.50×10-3 m salicylic acid, we need to calculate the number of moles of salicylic acid required. The formula for this is concentration x volume. Once we have the number of moles required, we can use the molar mass of salicylic acid to convert the number of moles to grams. This gives us the amount of salicylic acid needed to prepare the solution. In this case, we need 17.3 mg of salicylic acid to prepare the solution.

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for each solution tested determine which ion reacts with water ( ion hydrolyzed) and which ions didn't react with water

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In summary, to determine which ions react with water (ion hydrolyzed) and which ions don't react with water, we need to look at the nature of the cation and anion of the solute and determine if they are weak acids or weak bases. If they are, they will react with water to form their conjugate acid or base, respectively. Otherwise, they will not react with water.

To determine which ions react with water (ion hydrolyzed) and which ions don't react with water, we need to look at the dissociation of the solute in water. If the cation or anion of the solute is a weak acid or a weak base, it will react with water to form its conjugate acid or base, respectively. This reaction is called hydrolysis.
For example, if we have the solution of ammonium chloride (NH4Cl), the ammonium ion (NH4+) is a weak acid and will react with water to form hydronium ions (H3O+) and ammonia (NH3). The chloride ion (Cl-) is not a weak base and will not react with water.
NH4Cl + H2O ↔ NH4+ + Cl- + H3O+ + OH-
In another example, if we have the solution of sodium nitrate (NaNO3), both the cation (Na+) and the anion (NO3-) are neither a weak acid nor a weak base. Hence, they will not react with water.
NaNO3 + H2O ↔ Na+ + NO3- + H2O
In summary, to determine which ions react with water (ion hydrolyzed) and which ions don't react with water, we need to look at the nature of the cation and anion of the solute and determine if they are weak acids or weak bases.

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!!please hurry!!

Which of the following is a true statement?
(1 point)
Responses:

(A) When it is summer in the northern hemisphere, it is winter in the southern hemisphere.

(B) When it is day in the northern hemisphere, it is night in the southern hemisphere.

(C) When it is summer in the northern hemisphere, it is winter on the equator.

(D) When it is summer in the poles, it is winter on the equator.

Answers

The True statement is Option A. When it is summer in the northern hemisphere, it is winter in the southern hemisphere.

This is due to the Earth's tilt and its revolution around the Sun. The Earth is tilted at an angle of 23.5 degrees, which causes different parts of the planet to receive varying amounts of sunlight throughout the year. During the northern hemisphere's summer, the North Pole is tilted towards the Sun, which means it receives more direct sunlight, making it warmer. At the same time, the South Pole is tilted away from the Sun, making it colder, and hence it is winter in the southern hemisphere. This phenomenon is reversed during the northern hemisphere's winter, with the South Pole being tilted towards the Sun, and it is summer in the southern hemisphere.

Option (B) is incorrect because day and night occur due to the rotation of the Earth on its axis, and it is not related to the hemisphere's seasons. Option (C) is also incorrect because the equator does not experience winter or summer, but it does experience rainy and dry seasons. Option (D) is incorrect because the poles do not have distinct seasons, but they do experience periods of continuous daylight and darkness depending on their position relative to the Sun.

In conclusion, the correct statement is (A) When it is summer in the northern hemisphere, it is winter in the southern hemisphere, due to the Earth's tilt and revolution around the Sun.

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The enthalpy of solution is defined as ∆Hsolnv = ∆Hsolute + ∆Hsolvent + ∆Hmix. Each of the terms on the right side of the equation are either endothermic or exothermic. Which answer properly depicts this.

Answers

The terms ∆Hsolute, ∆Hsolvent, and ∆Hmix can be either endothermic or exothermic depending on the specific solute and solvent involved. Therefore, there is no single answer that properly depicts the signs of these terms.

The enthalpy of solution, which is the heat absorbed or released when a solute dissolves in a solvent, can be broken down into three component enthalpies:

Hsolute, which is the heat absorbed or released when the solute is dissolved in the solvent;

Hsolvent, which is the heat absorbed or released when the solvent is diluted by the solute; and

∆Hmix, which is the heat absorbed or released when the solute and solvent mix. Each of these three terms can be either endothermic or exothermic, depending on whether heat is absorbed or released during the process.

For example, if the solute dissolves in the solvent and releases heat, ∆Hsolute would be negative (exothermic), while if the solvent is diluted by the solute and absorbs heat, ∆Hsolvent would be positive (endothermic).

Therefore, the sign of each term in the equation depends on the specific solute and solvent involved and the conditions under which they are mixed.

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bitter taste is elicited by ________. bitter taste is elicited by ________. metal ions acids alkaloids hydrogen ions

Answers

The bitter taste is primarily elicited by alkaloids (option c). Alkaloids are a diverse group of naturally occurring organic compounds, mainly derived from plants, that contain nitrogen atoms.

Alkaloids are a class of compounds found in many plants that can also produce a bitter taste. These compounds are often associated with the medicinal properties of plants and are found in many herbal remedies and supplements.

They often have a bitter taste and are frequently found in foods and beverages such as coffee, tea, and certain vegetables. Some common examples of alkaloids include caffeine, nicotine, and quinine. Although metal ions, acids, and hydrogen ions can also contribute to taste perception, they are not the primary contributors to the bitter taste sensation.

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Which of the following ions contain a central atom with a formal charge? Select the correct answer below: O SCN- (C is the central atom) ОРО O CHỊ0 O CC+

Answers

The ions that contain a central atom with a formal charge are SCN- (with carbon, C, as the central atom) and CC+.

In SCN-, the central atom carbon (C) has a formal charge of +1, while the other atoms, sulfur (S) and nitrogen (N), have formal charges of 0 and -1, respectively.

In CC+, the central atom carbon (C) has a formal charge of +1.

Therefore, the correct answer is: SCN- (C is the central atom) and CC+.

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typically an oxygen atom with ____ covalent bond will have a formal charge of −1.

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Typically an oxygen atom with three covalent bonds will have a formal charge of −1 Participating in non-polar covalent bonds is oxygen.

This is due to the six valence electrons that oxygen atoms possess. This indicates that in order to reach octet configuration, it has 2 lone pairs and 2 unpaired electrons that are shared.

The two lone pairs on the oxygen atom in this chemistry are not shared with any other atoms. Instead, they are paired with the atom of oxygen. The oxygen atom has no formal charge. The atomic number of oxygen is 8, which is the total of its valence and inner shell electron counts.

A type of covalent bond known as a nonpolar covalent bond involves two atoms sharing the bonding electrons equally.

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Calculate the molar solubility and the solubility in g/L of each salt at 25 degreeC: PbF2 Ksp = 4.0 x 10-8 x 10 M g/L Ag2C03 Ksp = 8.1 x 10-12 x 10 M x 10 g/L Bi2S3 Ksp = 1.6 x 10-72 x 10 M x 10 g/L Enter all of your answers in scientific notation except the solubility of a .

Answers

The Molar solubility and the solubility of each salt at 25°C are: (a) PbF₂ : 4.41 x 10⁻⁵  g/L ; (b) Ag₂CO₃: 0.0398 g/L ; (c) Bi₂S₃ : 1.65 x 10⁻¹³ g/L

Let us consider X be the molar solubility of PbF₂.

Then, [Pb2+] = X and [F-] = 2X. Substituting into the Ksp expression and solving for x:

4.0 x 10⁻⁸ = X×(2X)²

X = 1.8 x 10⁻⁷ M

To convert to g/L, we need to multiply by the molar mass of PbF₂ (245.2 g/mol):

solubility = 1.8 x 10⁻⁷ × 245.2 = 4.41 x 10⁻⁵ g/L

(b) Ag₂CO₃ Ksp = [Ag⁺]²[CO₃²⁻]

Let x be the molar solubility of Ag₂CO₃. Then, [Ag+] = 2x and [CO₃²⁻] = x. Substituting into the Ksp expression and solving for x:

8.1 x 10⁻¹² = (2x)² × x

x = 1.2 x 10⁻⁴ M

To convert to g/L,

we will multiply by the molar mass of Ag₂CO₃ (331.8 g/mol):

Therefore, solubility = 1.2 x 10⁻⁴ × 331.8 = 0.0398 g/L

(c) Bi₂S₃ Ksp = [Bi³⁺]²[S²⁻]³

Let x be the molar solubility of Bi₂S₃. Then, [Bi³⁺] = 2x and [S²⁻] = 3x. Substituting into the Ksp expression and solving for x:

1.6 x 10⁻⁷² = (2x)²×(3x)³

x = 3.2 x 10⁻¹⁶

To convert to g/L, we need to multiply by the molar mass of Bi₂S₃ (514.2 g/mol):

solubility = 3.2 x 10⁻¹⁶ × 514.2 = 1.65 x 10⁻¹³ g/L

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the rate constant at 325 °c for the decomposition reaction c4h8 ⟶ 2c2h4 is 6.1 × 10−8 s −1, and the activation energy is 261 kj per mol of c4h8. determine the frequency factor for the reaction.

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The frequency factor for the decomposition reaction C4H8 ⟶ 2C2H4 with a rate constant of 6.1 × 10−8 s−1 at 325 °C and an activation energy of 261 kJ/mol is 2.3 × 10^12 s−1.

The frequency factor, denoted by A, can be calculated using the Arrhenius equation:

k = A * exp(-Ea/RT)

where k is the rate constant, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. We can first convert the temperature given in the question from Celsius to Kelvin:

T = 325 + 273.15 = 598.15 K

Now, we can plug in the values given in the question:

6.1 × 10−8 s−1 = A * exp(-261000 J/mol / (8.314 J/mol*K * 598.15 K))

Simplifying the right side of the equation:

6.1 × 10−8 s−1 = A * exp(-43.58)

Solving for A:

A = 6.1 × 10−8 s−1 / exp(-43.58)

A = 2.3 × 10^12 s−1

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How many moles of potassium chloride are needed to react with 9. 27 moles of


oxygen gas?


2KCI (s) + 302 (g) - — 2KCIO3 (s)

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To determine the number of moles of potassium chloride (KCl) required to react with 9.27 moles of oxygen gas ( O_{2}), we need to use the stoichiometry of the balanced chemical equation. The balanced equation shows that 2 moles of potassium chloride react with 3 moles of oxygen gas to produce 2 moles of potassium chlorate ([tex]KClO_{3}[/tex]).

According to the stoichiometry of the balanced chemical equation, 2 moles of potassium chloride react with 3 moles of oxygen gas to produce 2 moles of potassium chlorate. Therefore, we can set up a ratio based on this stoichiometry:

2 moles KCl / 3 moles O_{2}= x moles KCl / 9.27 moles O_{2}

Solving for x, we can find the number of moles of potassium chloride required:

x = (2 moles KCl / 3 moles O_{2}) * 9.27 moles [tex]O_{2}[/tex]

x = 6.18 moles KCl

Therefore, 6.18 moles of potassium chloride are needed to react with 9.27 moles of oxygen gas. The stoichiometry of the balanced equation allows us to determine the appropriate amounts of reactants required for the given reaction.

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A physican ordered the ptosin infusion to run at 16ml/min. the pharamacy set up 10 units of ptosin in 500 ml of d5lr. you would set your pump at what ml/hr?

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The pump should be set at 19.2 ml/hr to deliver the Pitocin infusion at a

rate of 16 ml/min.

We need to calculate the infusion rate in ml/hr, given that the physician

has ordered the Pitocin infusion to run at 16 ml/min, and the pharmacy

has set up 10 units of Pitocin in 500 ml of D5LR.

First, we need to convert the infusion rate from ml/min to ml/hr:

16 ml/min x 60 min/hr = 960 ml/hr

Next, we need to calculate the infusion rate of the Pitocin solution. We

know that the solution contains 10 units of Pitocin in 500 ml of D5LR, so

the concentration of Pitocin in the solution is:

10 units/500 ml = 0.02 units/ml

Finally, we can use the concentration and the infusion rate to calculate

the infusion rate of Pitocin:

0.02 units/ml x 960 ml/hr = 19.2 units/hr

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given the information a bc⟶2d⟶dδ∘δ∘=−723.0 kjδ∘=324.0 j/k=547.0 kjδ∘=−225.0 j/k calculate δ∘ at 298 k for the reaction a b⟶2c

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The standard entropy change for the reaction a b ⟶ 2c is -0.5 kJ/K/mol at 298 K.

The standard enthalpy change for the reaction a b ⟶ 2d is -723.0 kJ/mol, and the standard enthalpy change for the reaction 2d ⟶ d is -324.0 J/K/mol. The standard entropy change for the reaction d ⟶ δ is -547.0 J/K/mol, and the standard entropy change for the reaction a + b ⟶ 2c is unknown.

To find the standard enthalpy change for the reaction a b ⟶ 2c, we can use Hess's Law, which states that the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps. We can write the overall reaction as:

a b ⟶ 2c + 2d ⟶ 2δ

The enthalpy change for this reaction can be calculated as:

ΔH° = 2ΔH°(d ⟶ δ) + 2ΔH°(a b ⟶ 2d) - ΔH°(2c ⟶ 2δ)ΔH° = 2(-324.0 J/K/mol) + 2(-723.0 kJ/mol) - 0ΔH° = -1764.0 kJ/mol

Therefore, the standard enthalpy change for the reaction a b ⟶ 2c is -1764.0 kJ/mol.

To find the standard entropy change for the reaction a b ⟶ 2c, we can use the equation:

ΔG° = ΔH° - TΔS°

At 298 K, we have:

ΔG° = -1764.0 kJ/mol - (298 K)(-0.547 kJ/K/mol)ΔG° = -1614.9 kJ/mol

We can rearrange this equation to solve for ΔS°:

ΔS° = (ΔH° - ΔG°) / TΔS° = (-1764.0 kJ/mol - (-1614.9 kJ/mol)) / 298 KΔS° = -0.5 kJ/K/mol

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What volume of the oxygen gas, measured at 27 degree C and 0.987 atm, is produced from the decomposition of 67.5 g of HgO(s)? 2HgO(s) rightarrow 2 Hg(1) + O_2(g) A. 7.77 L B. 6.98 L C. 3.89 LD. 3.49 L
Mlar mass HgO = 216.59

Answers

The volume of oxygen gas produced is 6.98 L

So, the correct answer is B.

To calculate the volume of oxygen gas produced, we first need to determine the moles of HgO decomposed.

Using the given mass (67.5 g) and molar mass of HgO (216.59 g/mol), we find:

moles of HgO = 67.5 g / 216.59 g/mol = 0.3119 mol

From the balanced equation, 2 moles of HgO produce 1 mole of O₂.

So, moles of O₂ produced = 0.3119 mol HgO * (1 mol O₂ / 2 mol HgO) = 0.1559 mol O₂.

Next, we'll use the ideal gas law (PV=nRT) to find the volume of O₂:

V = nRT/P = (0.1559 mol)(0.0821 L⋅atm/mol⋅K)(27°C + 273.15K) / 0.987 atm = 6.98 L, which is option B.

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how many electrons are exchanged in total when the reaction cr2o72- so32- --> cr3 so42- is run?

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A total of 6 electrons are exchanged in the reaction.

In the redox reaction Cr2O7^2- + SO3^2- → Cr^3+ + SO4^2-, a total of 6 electrons are exchanged. The Cr2O7^2- ion is reduced to two Cr^3+ ions, each gaining 3 electrons, and the SO3^2- ion is oxidized to SO4^2-, losing 2 electrons. The balanced half-reactions are:

Cr2O7^2- + 14H^+ + 6e- → 2Cr^3+ + 7H2O (reduction)
2SO3^2- → 2SO4^2- + 2e- (oxidation)

To balance the electrons exchanged, multiply the oxidation half-reaction by 3:

6SO3^2- → 6SO4^2- + 6e-

Thus, a total of 6 electrons are exchanged in the reaction.

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what current (in a) is required to plate out 1.22 g of nickel from a solution of ni2 in 3.0 hour?

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A current of approximately 12.7 mA is required to plate out 1.22 g of nickel from a solution of Ni2+ in 3.0 hours.

To calculate the current required to plate out 1.22 g of nickel from a solution of Ni2+ in 3.0 hours, we need to use Faraday's Law of Electrolysis.

The equation for Faraday's Law is:
Amount of substance plated = (Current x Time x Atomic weight) / (Charge per mole of electrons)

In this case, the amount of substance plated is 1.22 g of nickel. The atomic weight of nickel is 58.69 g/mol. The charge per mole of electrons is 2 (since Ni2+ has a charge of 2+).

So, we can rearrange the equation to solve for the current:
Current = (Amount of substance plated x Charge per mole of electrons) / (Time x Atomic weight)

Plugging in the values:
Current = (1.22 g x 2) / (3.0 hours x 58.69 g/mol)
Current = 0.0127 A or 12.7 mA (rounded to two significant figures)

Therefore, a current of approximately 12.7 mA is required to plate out 1.22 g of nickel from a solution of Ni2+ in 3.0 hours.

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gastric (stomach) secretions are one of the only solutions in the body that are not buffered because

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The gastric secretions are not buffered because they need to maintain their acidic ph in order to properly digest food. The stomach secretes hydrochloric acid and other enzymes to break down food.

The acidic environment is necessary for the enzymes to function properly and for the stomach to effectively digest proteins. Buffering the acid would interfere with this process and potentially cause digestive issues. Therefore, the body has evolved to allow gastric secretions to remain unbuffered.

Most body fluids are buffered to maintain a stable pH to prevent damage to cells and tissues. However, in the case of gastric secretions, the low pH high acidity is necessary for effective digestion. If gastric secretions were buffered, the stomach would not be able to efficiently break down food and initiate the digestive process.

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the formula for the oxalate ion is c2o42− . predict the formula for oxalic acid.

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The formula for the oxalate ion is C2O4²⁻. To predict the formula for oxalic acid, we need to consider that an acid is formed when a hydrogen ion (H⁺) combines with an anion. In this case, the anion is the oxalate ion.

Oxalic acid is a dibasic acid, which means it can donate two protons (H⁺) to form two salt ions. As the oxalate ion has a 2- charge, it will require two hydrogen ions to neutralize this charge and form the corresponding acid. So, each oxalate ion will combine with two hydrogen ions to create a neutral compound.

With this information, we can now predict the formula for oxalic acid. Combining two hydrogen ions (H⁺) with the oxalate ion (C2O4²⁻) results in the chemical formula H₂C₂O₄. This is the formula for oxalic acid, which is a weak organic acid found in various natural sources, such as vegetables and fruits. It is also used in various industrial applications as a cleaning agent, rust remover, and bleach.

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