PLEASE HELP!!!


A sample of nitrogen at 20.0 °C has a volume of 588 mL at a pressure of 924 mmHg. To change the volume to 1008 mL, what must be the new pressure?

Answer:

(i) Change in colour (ii) Change in temperature  (iii) Formation of precipitate

Explanation:

(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) →  BaSO4(s) +2NaCl(aq) BaSO4(s)

Answer:

B bone marrow

Explanation:

I hope it will help

Answer:

C,A,B,D

Explanation:

That's the correct order, hope this helps

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:[tex]X^{m+}Y^{n-}\rightarrow X_nY_m[/tex]

Thus, for all the given combinations, we obtain:

- Y⁻

[tex]X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3[/tex]

- Y²⁻

[tex]X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3[/tex]

- Y³⁻

[tex]X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY[/tex]

Best regards!

Answer:

i think the answer is b temperature

Answer:

no

Explanation:

Answer:

%N = 25.94%

%O = 74.06%

Explanation:

Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅

We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.

m(N): 2 × 14.01 g = 28.02 g

Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅

We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.

m(O): 5 × 16.00 g = 80.00 g

Step 3: Calculate the mass of 1 mole of N₂O₅

We will sum the masses of N and O.

m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g

Step 4: Calculate the percent composition  of N₂O₅

We will use the following expression.

%Element = m(Element)/m(Compound) × 100%

%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%

%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%

Answer:

Terbium

Explanation:

Answer:

[tex]\Delta S_{sys}=0.020kJ=20J[/tex]

Explanation:

Hello!

In this case, given the required energy to vaporize 1.00 mol of ethanol as the enthalpy of vaporization:

[tex]\Delta H_{vap}=\frac{38.6kJ}{1.00mol}=38.6kJ/mol[/tex]

We can compute the entropy of the system for the vaporization of 8.00 g of ethanol, by first computing the moles:

[tex]n_{et}=8.00g*\frac{1mol}{46.07g} =0.174mol[/tex]

And then setting up the following expression:

[tex]\Delta S_{sys}=\frac{n_{et}*\Delta H_{vap}}{T}[/tex]

Whereas the temperature is in kelvins; thus, we obtain:

[tex]\Delta S_{sys}=\frac{0.174mol*38.6\frac{kJ}{mol} }{79.6+273.15K}\\\\\Delta S_{sys}=0.020kJ=20J[/tex]

Best regards!

The entropy of the system ([tex]\Delta S_{sys}[/tex]) for the vaporization of 8.00 grams of ethanol at 79.6 °C is 0.0190 kilojoules.

Given the following data:

Conversion:

Temperature = [tex]79.6 + 273 = 352.6 \;K[/tex]

To find the entropy of the system ([tex]\Delta S_{sys}[/tex]) for the vaporization of 8.00 grams of ethanol at 79.6 °C:

First of all, we would determine the heat of vaporization.

[tex]Heat\; of \;vaporization = \frac{Energy}{moles} \\\\Heat\; of \;vaporization = \frac{38.6}{1}[/tex]

Heat of vaporization = 38.6 kJ/mol.

Next, we would determine the number of moles in 8.00 grams of ethanol:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{8}{46.07}\\\\Number\;of\;moles = 0.1737 \;moles[/tex]

Mathematically, the entropy of a system ([tex]\Delta S_{sys}[/tex]) is given by the formula:

[tex]\Delta S_{sys} = \frac{n\Delta H }{T}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\Delta S_{sys} = \frac{0.1737 \times 38.6 }{352.6}\\\\\Delta S_{sys} = \frac{6.7048 }{352.6}\\\\\Delta S_{sys} = 0.0190 \;kJ[/tex]

Read more: https://brainly.com/question/6366973

Answer:

d. Although the hydrolysis of ATP is thermodynamically favorable, without a catalyst the reaction occurs at a very slow rate because it has a large activation energy.

Explanation:

Hello!

In this case, since negative Gibbs free energies of reaction stand for thermodynamically favored processes, we can immediately rule out choices a. and b.

Moreover, since the reaction is slow without the presence of a catalyst, which the context of biochemistry is an enzyme, we infer that correct choice is d. Although the hydrolysis of ATP is thermodynamically favorable, without a catalyst the reaction occurs at a very slow rate because it has a large activation energy because the higher the activation energy the slower the reaction according to the Arrhenius equation.

Best regards!

Answer:

Convection

Explanation:

Convection is the circular motion that happens when warmer air or liquid — which has faster moving molecules, making it less dense — rises, while the cooler air or liquid drops down.

Same as in the case of sea breeze where the the lighter wind in surface of land rises up and the cooler wind from the sea comes to fill up that place. This is an example of convection.

Answer:

I promise it is the 2nd one

Explanation:

The following diagram shows a stage of a cell during mitosis. Two identical circles are joined in the middle and they have the nucleolus inside the nucleus in the center of the circles.

Answer:

The Second One

Explanation:I did the test and got this question correct

Pls comment :D

Answer:

Cellular respiration

Explanation:

Answer:

98.68%

Explanation:

Step 1: Write the balanced equation

KBr + AgNO₃ ⇒ AgBr + KNO₃

Step 2: Calculate the moles corresponding to 814.5 mg (0.8145 g) of AgBr

The molar mass of AgBr is 187.77 g/mol.

0.8145 g × 1 mol/187.77 g = 4.338 × 10⁻³ mol

Step 3: Calculate the moles of KBr needed to produce 4.338 × 10⁻³ moles of AgBr

The molar ratio of KBr to AgBr is 1:1. The moles of KBr needed are 1/1 × 4.338 × 10⁻³ mol = 4.338 × 10⁻³ mol.

Step 4: Calculate the pure mass corresponding to 4.338 × 10⁻³ moles of KBr

The molar mass of KBr is 119.00 g/mol.

4.338 × 10⁻³ mol × 119.00 g/mol = 0.5162 g

Step 5: Calculate the purity of KBr

0.5162 g of KBr are in a 0.5231 g-sample. The purity of KBr is:

P = 0.5162 g/0.5231 g × 100% = 98.68%

Ammonia chemical formula : NH3

Ammonia molar mass : 14 + (1×3) = 17

Mass = Mol × mm

= 3.5mol × 17 = 59.5g

See pic for formula

nvm i think your question is incomplete. pls report my answer i thought u are asking the mass of ammonia

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

The percentage composition of 41K isotope of potassium according to the given data is 6.7302 percent.

Given that the atomic masses of the potassium isotopes are:

mass of 39K = 38.9637amu

mass of 40K = 39.9639amu

mass of 41K = 40.9618amu

and the percentages are:

percentage of 39K = 93.2581%

percentage of 40K =  0.0117%

percentage of 41K = x

Given average mass of potassium = 39.0983amu

sum (mass of isotopes × percentage abundance) =  average mass

38.9637 × 93.2581% +  39.9639 ×  0.0117%  + 40.9618 × x = 39.0983

x = 6.7302%

So the percentage composition of 41K is 6.7302%

Learn more about composition of isotopes:

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Answer is B

A mixture means two compounds mixed together but not atomically. For example, sand and water solution. A water with sand mixed inside is a mixture because the sand can be separated by physical means.

Water is H2O. This means it is a compound, because hydrogen and oxygen is bonded together atomically. Hydrogen however is an element, because you can't seperate it anymore (You could seperate it in the sense of protons and neutrons, but the definition of element is something you can see on the periodic table)

Compound means more than one element bonded together.

The sample must be a compound because it couldn't be seperated by physical means but can be seperated by chemical means into three pure substances (synonym of element)

Answer:

Option D and Option B

When comparing the two elements K and Ge , the more metallic element is_____K_____ based on periodic trends alone.

When comparing the two elements Sb and Pb , the more metallic element is_____Pb______ based on periodic

Explanation:

The metallic characteristic increases when we move down a column in a periodic table or when we move left in the row.

Potassium and Germanium are located on the same row, but germanium lies on the right side of potassium. Thus, potassium (K) is more metallic than Germanium (Ge)

While Lead (Pb) lies to left of Sb in the adjacent column and is also lies below Sb. Hence Pb is more metallic than Sb

Answer:

16.5g of CO₂ could be produced

Explanation:

The combustion of hexane occurs as follows:

C₆H₁₄(l) + 19/2O₂ → 6CO₂ + 7H₂O

Where 1 mole of hexane reacts with 19/2 moles of O₂.

To solve this question we need to find the moles of each reactant in order to find limiting reactant. The moles of the limiting reactant will determine the moles of CO₂ produced:

Moles C₆H₁₄ -Molar mass: 86.18g/mol-:

17.2g hexane * (1mol / 86.18g) = 0.200 moles hexane

Moles O₂ -Molar mass: 32g/mol-:

19g O₂ * (1mol / 32g) = 0.594 moles oxygen.

For a complete reaction of 0.594 moles of oxygen are required:

0.594 moles O₂ * (1mol C₆H₁₄ / 19/2 moles O₂) = 0.0625 moles C₆H₁₄.

As there are 0.200 moles of hexane, hexane is the excess reactant and oxygen the limiting reactant.

The moles of CO₂ produced assuming a yield of 100% -All moles of oxygen react producing carbon dioxide.:

0.594 moles O₂ * (6mol CO₂ / 19/2 moles O₂) = 0.375 moles of CO₂ could be produced. The mass is:

0.375 moles of CO₂ * (44.01g / mol) =  

Answer: a)  The maximum mass of sulfur trioxide that can be formed is 31.4 grams

b) The FORMULA for the limiting reagent is [tex]O_2[/tex]

c) Mass of excess reagent remains is 4.8 grams

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles[/tex]

[tex]\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles[/tex]

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]  

According to stoichiometry :

1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]

Thus 0.196 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_2[/tex]

Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]SO_2[/tex] is the excess reagent as (0.467-0.392) = 0.075 moles or [tex]0.075mol\times 64g/mol=4.8g[/tex] are left.  

As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]SO_3[/tex]

Thus 0.196 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_3[/tex]

Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g[/tex]

Thus 31.4 g of [tex]SO_3[/tex] will be produced from the given masses of both reactants.

690 g AgCl

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂

↓

[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

[Given] 2.4 mol ZnCl₂

[Solve] x g AgCl

Step 2: Identify Conversions

[RxN] 1 mol ZnCl₂ → 2 mol AgCl

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

Step 3: Stoich

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

687.936 g AgCl ≈ 690 g AgCl

Answer:

There are 0.5 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number  of particles.

To calculate the moles, we use the equation:

Thus there are 0.5 moles

Explanation:

Answer:

i think its Yes. The bubbles contain hydrogen and oxygen that separated from the water but i could be wrong so i would just wait to see if anybody else says anything

Explanation:

Answer:

Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)

Explanation:

Decide whether a chemical reaction happens in either of the following situations.

- A reaction happens. The type of this reaction is the single replacement reaction.

If a reaction does happen, write the chemical equation for it.

Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)

Carbon-14 is a radioisotope of carbon that decays following first-order kinetics. There are four values of interest in this problem: the "normal" (or original) amount of carbon-14 for a jawbone ([tex]\mathrm{N_0}[/tex]), the actual amount of carbon-14 in a jawbone ([tex]\mathrm{N}[/tex]), the half-life of carbon-14 ([tex]\mathrm{t_{1/2}}[/tex]), and the actual time elapsed ([tex]\mathrm{t}[/tex]) from the original time. There is an equation that ties all these values in together,

[tex]N= N_0 e^{-kt}[/tex]

where k is the rate constant, which, for first-order decay, is related to the half-life by

[tex]k = \dfrac{\ln 2}{ t_{1/2} }.[/tex]

What you want to find here is the time elapsed (t). So, you can substitute the latter equation for k into the k in the former equation to get

[tex]N= N_0 e^{\frac{-\ln 2 \;t}{t_{1/2}}.[/tex]

Rearranging to solve for t, the equation becomes

[tex]t = \left(\dfrac{\ln \dfrac{N_0}{N}}{\ln 2} \right) t_{1/2}.[/tex]

You are given all three of the values necessary to solve for t: The normal amount of carbon-14 is 200 mg; the actual amount of carbon-14 in the sample is 50 mg; and the half-life of carbon-14 is 5730 years. Plugging them into the above equation, we get

[tex]t = \left(\dfrac{\ln \dfrac{200 \text{ mg}}{50 \text{ mg}}}{\ln 2} \right) \left(5730 \text{ years} \right) = 11460 \text{ years}.[/tex]

So the jawbone found is 11460 years old (or 11000 if accounting for sig figs).

Answer:the answer is a c and e

Explanation:

Answer:

Explanation:

From the information given:

The Chemical equation is:

[tex]SO_2Cl_{2(g)} \iff SO_{2(g)} + Cl_{2(g)}\\[/tex]

since 43.6% of the initial concentration remains at equilibrium

Then; the amount of [tex]SO_2Cl_2[/tex] that is being reacted is:

= 0.543 × (100 -43.6)%

= 0.306 M

The ICE table can be computed as follows:

           [tex]SO_2Cl_2[/tex]               ⇔           [tex]SO_{2(g)[/tex]           +             [tex]Cl_{2(g)[/tex]

I            0.543                                    0                           0

C          0.306                                 +0.306                     0.306

E           0.237                                  0.306                      0.306

[tex]K_c = \dfrac{[SO_2] [Cl_{2}]}{[SO_2Cl_2]}[/tex]

[tex]K_c = \dfrac{0.306 \times 0.306}{0.237}[/tex]

[tex]K_c = 0.995[/tex]

Thus; the concentration at equilibrium for the species are:

[tex]SO_2Cl_2[/tex]  = 0.237 M

[tex]SO_{(2g)[/tex] = 0.306 M

[tex]Cl_{2(g)[/tex] = 0.306 M

Answer:

it is directly proportional to the temperature of the gas

Explanation: