Compound A has a partition coefficient (K) of 7 when comparing its solubility in CH2Cl2 to water ( K=11, [solubility of A in g/ml in CH2Cl2] divided by [solubility of A in g/ml in H2O]). If we take 17.0 g of A and partition it thoroughly between 150 ml of CH2Cl2 and 100 ml of water, what is the equation which will tell us how much of A (which is represented by x) is in the water layer after this partitioning ?

Answer: The mass of Cu produced is 4.88 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of aluminum = 2.98 g

Molar mass of aluminum = 27 g/mol

Plugging values in equation 1:

[tex]\text{Moles of aluminum}=\frac{2.98g}{27g/mol}=0.1104 mol[/tex]

The given chemical equation follows:

[tex]2Al(s)+3CuSO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3Cu(s)[/tex]

By the stoichiometry of the reaction:

If 2 moles of aluminum produces 3 moles of Cu

So, 0.1104 moles aluminium will produce = [tex]\frac{3}{2}\times 0.1104=0.1656mol[/tex] of Cu

Molar mass of Cu = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Mass of Cu}=(0.1656mol\times 63.5g/mol)=10.516g[/tex]

The percent yield of a reaction is calculated by using an equation:

[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(2)

Given values:

% yield of product = 46.4 %

Theoretical value of the product = 10.516 g

Plugging values in equation 2, we get:

[tex]46.4=\frac{\text{Actual value of Cu}}{10.516g}\times 100\\\\\text{Actual value of Cu}=\frac{46.4\times 10.516}{100}\\\\\text{Actual value of Cu}=4.88g[/tex]

Hence, the mass of Cu produced is 4.88 g

I think it's 9 ................'

Answer:

C₄H₄N₂

Explanation:

Given that:

M+ = 80.

It implies that the number of nitrogen present in the molecule must also be even according to the Nitrogen rule.

So from the Formula CHN, the nitrogen will have to be 2 because if we make use of 4, it will exceed the given M+ which is 80.

∴

C₄ = 4 × 12 = 48

H₄  = 4 × 1   = 4

N₂ =  2 × 14 = 28      

                      80      

As such, the molecular formula of the compound is C₄H₄N₂

Answer:

[tex]m_{HCl}=36.1gHCl[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

[tex]n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2[/tex]

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

[tex]m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl[/tex]

Regards!

Answer:

yes the experimental result is the expected result .

Explanation:

When Titrating acetate buffer with acid the PH will decrease gradually from a more neutral PH to a more acidic level and this is because buffer solutions are prepared with weak acids and its conjugate base.

The results gotten from the continuous addition of base NaOH to the acetate buffer is the expected result because the base is been absorbed by the buffer solution and it is converted to a conjugate base of the buffer solution which will gradually increase the PH level of the solution as more conjugate base is formed due to the addition of more NaOH.  

Answer:

If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.

Explanation:

The balanced reaction is:

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 1 mole of C₅H₁₂ produces 5 moles of CO₂, then 7.9 moles of C₅H₁₂ will produce how many moles of CO₂?

[tex]amount of moles of CO_{2} =\frac{7.9 moles of C_{5}H_{12}*5 moles of CO_{2} }{1 mole of C_{5}H_{12} }[/tex]

amount of moles of CO₂= 39.5 moles

If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

Answer:

h2o

Explanation:

Explanation:

Iron atom is been oxidised as it losses 2 electron to form 2 + ion.

Answer:

0050 hours

Explanation:

The size of the drops is regulated by types of tubing in microdrip tubing 1cc(1ml ) forms 60 gtt.

Gtt in 500 cc D5W is 500×60 gtt.= 30000 gtt.

Time required to infuse whole fluid with the rate of 60gtt /min = 30000gtt÷60gtt/min = 500minute.

= 8hours 20 minute.

Fluid infusion started at 1630 hours after and it ends after 8 hours 30 minutes .

Therefore, infusion will be completed at 1630 hours +0830 hours.= 0050 hours.

Answer:

A physical change is a change in form.

A Chemical change is a change in materials.

Explanation:

Example of a physical change would be an ice cube meting.

Example of a chemical change would be mixing food coloring into a cup of water.

Answer:

Option E

Explanation:

From the question we are told that:

Amount of sand in percentage [tex]s_p=45%[/tex]

Sample size[tex]n=120g[/tex]

Note:After being dumped in the river repeatedly the sugar melts away leaving behind the insoluble sand

Generally the equation for Amount of sand content is mathematically given by

 [tex]X=n*s_p[/tex]

 [tex]X=120*\frac{45}{100}[/tex]

 [tex]X=54g[/tex]

Therefore

After drying, the mass of the contents of the bag equals

 [tex]X=54g[/tex]

Option E

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

Answer:

As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.

I hope this will help you if not soo sorry :)

Explanation:

strong electrolyte- Nacl HCL NAOH

weak electrolyte- c12H22O, NH3

NaCl,HCl and NaOH  are strong electrolytes while ammonia is a weak electrolyte and sucrose is a non-electrolyte.

It is a solution which consists of ions which are electrically conducting as a result of movement of ions.Class of electrolytes include most soluble salts,acids and bases which are dissolved in a polar solvent.On dissolution, they separate into the constituent ions.

There are 3 classes according to the nature of substance which results upon dissolution:

1) Strong electrolytes- Substances which on dissolution in a medium dissociate completely are strong electrolytes. eg: NaCl,HCl

2) Weak electrolytes-  Substances which on dissolution in a medium dissociate partially are weak electrolytes. eg: NH₃

3)Non-electrolytes- Substances which do not dissociate on dissolution are non-electrolytes. eg: sucrose

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Answer:

D.) ±0.01 cm?​

Explanation:

Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.

The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.

Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.

So, we have our uncertainty as 0.01 cm.

So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.

So, the answer is D. which is ± 0.01 cm.

AAnswer:A

Explanation:

Answer:

That means Cu2O is limiting reagent and C is excess reagent

Explanation:

Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.

To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.

Moles Cu2O -Molar mass: 143.09 g/mol-

114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O

Moles C -Molar mass: 12.01g/mol-

11.1g C * (1mol / 12.01g) = 0.924 moles C

Answer:

Some bacteria like rhizobium and blue green algae are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called biological nitrogen fixers.

Answer:

Diazotrophic bacteria or Cyanobacteria

Answer:

Compare the solubility of silver chloride in each of the following aqueous solutions:

a. 0.10 M AgNO3 More soluble than in pure water.

b. 0.10 M NaCI Similar solubility as in pure water

c. 0.10 M KNO3 Less soluble than in pure water.

d. 0.10 M NH4CH3COO

Explanation:

This is based on common ion effect.

According to common ion effect, the solubility of a sparingly soluble salt decreases in a solution containing common ion to it.

The solubility of AgCl(s) is shown below:

[tex]AgCl(s) <=> Ag^{+}(aq)+Cl^-(aq)[/tex]

So, when it is placed in:

a. 0.10 M AgNO3

Due to common ion effect Ag+, its solubility is less in this solution than in pure water.

b. 0.10 M NaCI :

Due to common ion effect Cl-, its solubility is less in this solution than in pure water.

c. 0.10 M KNO3 :

In this solution there is no presence of common ion.

So, the solubility of AgCl in this solution is similar to that of pure water.

d. 0.10 M NH4CH3COO:

In this solution, AgCl forms a precipitate.

So, the solubility of AgCl is more in this solution compared to pure water.

Answer:

história phkfk

Explanation:

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Answer:

0.51

Explanation:

Given the Nernst equation;

E= E° - 0.0592/n logQ

E= 355 mV or 0.355 V

E° = 0.34 - 0= 0.34 V

n= 2(two electrons were transferred in the process)

Equation of the reaction;

H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)

Substituting values;

0.355 = 0.34 - 0.0592/2 log([H^+]/1)

0.355 - 0.34 = - 0.0296 log [H^+]

0.015/-0.0296 = log [H^+]

Antilog (-0.5068) = [H^+]

[H^+] = 0.311 M

pH = -log[H^+]

pH= - log(0.311 M)

pH = 0.51

The potential difference between the half cell of the electrochemical cell is called cell potential. The pH of the solution at 355 mV will be 0.51.

An electrochemical cell generates electricity from the redox chemical reactions occurring inside the cell.

The balanced chemical reaction is shown as,

[tex]\rm H_{2}(g) + Cu^{2+}(aq) \rightarrow 2H^{+}(aq) + Cu(s)[/tex]

Using the Nernst equation:

[tex]\rm E= E^{\circ} - \dfrac{0.0592}{n }logQ[/tex]

Given,

E = 0.355 V

E° = 0.34 V

n = 2

Substituting values in the above equation:

[tex]\begin{aligned} 0.355 &= 0.34 - \dfrac{0.0592}{2} \;\rm log(\dfrac{[H^{+}]}{1})\\\\0.355 - 0.34 &= - 0.0296 \rm \; log [H^{+}]\\\\\dfrac{0.015}{-0.0296} &= \rm \; log [H^{+}]\end{aligned}[/tex]

Solving further,

[tex]\begin{aligned} \rm Antilog (-0.5068)& = \rm [H^{+}]\\\\\rm [H^{+}] &= 0.311 \;\rm M \end{aligned}[/tex]

The pH of the solution is calculated as:

[tex]\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= \rm - log(0.311\; M)\\\\&= 0.51\end{aligned}[/tex]

Therefore, 0.51 is the pH of the solution.

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Answer:

my define it will be turst me is c

Answer:

Newton's first and third law of Motion

Explanation:

The laws applying in the example Newton's first and third laws of Motion.

Answer:

It Newtons first, second, and third laws

Explanation:

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

Answer:

See explanation

Explanation:

If we look at the question closely, we will notice that the metal in question must be aluminum.

When aluminum is treated with sodium hydroxide, a precipitate, aluminium hydroxide is formed as follows;

Al(s) + 3NaOH(aq) ---> Al(OH)3(s) + 3Na(s)

In excess sodium hydroxide, the precipitate dissolves as follows;

Al(OH)3(s) + NaOH(aq) ----> [NaAlOH4]^-(aq)

The complex formed is sodium aluminum tetrahydroxo aluminate III.

The reaction of aluminum faith dilute hydrochloric acid occurs as follows to yield aluminum chloride;

2Al(s) + 6HCl(aq) ----> 2AlCl3(aq) + 3H2(g)

When aluminum metal is heated strongly, it yields aluminum oxide;

2Al(s) + 3O2(g) ---> Al2O3(s)

Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

Based on the equilibrium:

Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺

kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

Answer and Explanation:

We have to identify the anion (negatively charged ion) and the positive ion to form each compound. The sum of the positive and negative charges will be equal to 0 for a neutral compound.

Chloride: the anion is Cl⁻ (1 negative charge).

Magnesium (Mg²⁺) + Chloride (Cl⁻) : MgCl₂

Sodium (Na⁺) + Chloride (Cl-): NaCl

Zinc (Zn²⁺) + Chloride (Cl-): ZnCl₂

Lithium (Li⁺) + Chloride (Cl-) : LiCl

Lead(II) (Pb²⁺) + Chloride (Cl⁻): PbCl₂

Calcium (Ca²⁺) + Chloride (Cl⁻): CaCl₂

Iron(II) (Fe²⁺) + Chloride (Cl⁻): FeCl₂

Iron(III) (Fe³⁺) + Chloride (Cl⁻): FeCl₃

Potassium (K⁺) + Chloride (Cl): KCl

Nitrate: the anion is NO₃⁻ (1 negative charge).

Magnesium (Mg²⁺) + Nitrate (NO₃⁻) : Mg(NO₃)₂

Sodium (Na⁺) + Nitrate (NO₃⁻): NaNO₃

Zinc (Zn²⁺) + Nitrate (NO₃⁻): Zn(NO₃)₂

Lithium (Li⁺) + Nitrate (NO₃⁻) : LiNO₃

Lead(II) (Pb²⁺) + Nitrate (NO₃⁻): Pb(NO₃)₂

Calcium (Ca²⁺) + Nitrate (NO₃⁻): Ca(NO₃)₂

Iron(II) (Fe²⁺) + Nitrate (NO₃⁻): Fe(NO₃)₂

Iron(III) (Fe³⁺) + Nitrate (NO₃⁻): Fe(NO₃)₃

Potassium (K⁺) + Nitrate (NO₃⁻): KNO₃

Sulphate: SO₄²⁻ (2 negative charges)

Magnesium (Mg²⁺) + Sulphate (SO₄²⁻) : MgSO₄

Sodium (Na⁺) + Sulphate (SO₄²⁻): Na₂SO₄

Zinc (Zn²⁺) + Sulphate (SO₄²⁻): ZnSO₄

Lithium (Li⁺) + Sulphate (SO₄²⁻) : Li₂SO₄

Lead(II) (Pb²⁺) + Sulphate (SO₄²⁻): PbSO₄

Calcium (Ca²⁺) + Sulphate (SO₄²⁻): CaSO₄

Iron(II) (Fe²⁺) + Sulphate (SO₄²⁻): FeSO₄

Iron(III) (Fe³⁺) + Sulphate (SO₄²⁻): Fe₂(SO₄)₃

Potassium (K⁺) + Sulphate (SO₄²⁻): K₂SO₄

Carbonate: CO₃²⁻ (2 negative charges)

Magnesium (Mg²⁺) + Carbonate (CO₃²⁻) : MgCO₃

Sodium (Na⁺) + Carbonate (CO₃²⁻): Na₂CO₃

Zinc (Zn²⁺) + Carbonate (CO₃²⁻): ZnCO₃

Lithium (Li⁺) + Carbonate (CO₃²⁻): Li₂CO₃

Lead(II) (Pb²⁺) + Carbonate (CO₃²⁻): PbCO₃

Calcium (Ca²⁺) + Carbonate (CO₃²⁻): CaCO₃

Iron(II) (Fe²⁺) + Carbonate (CO₃²⁻): FeCO₃

Iron(III) (Fe³⁺) + Carbonate (CO₃²⁻): Fe₂(CO₃)₃

Potassium (K⁺) + Carbonate (CO₃²⁻): K₂CO₃

Hydroxide: OH⁻ (1 negative charge)

Magnesium (Mg²⁺) + Hydroxide (OH⁻): Mg(OH)₂

Sodium (Na⁺) + Hydroxide (OH⁻): NaOH

Zinc (Zn²⁺) + Hydroxide (OH⁻): Zn(OH)₂

Lithium (Li⁺) + Hydroxide (OH⁻): LiOH

Lead(II) (Pb²⁺) + Hydroxide (OH⁻): Pb(OH)₂

Calcium (Ca²⁺) + Hydroxide (OH⁻): Ca(OH)₂

Iron(II) (Fe²⁺) + Hydroxide (OH⁻): Fe(OH)₂

Iron(III) (Fe³⁺) + Hydroxide (OH⁻): Fe(OH)₃

Potassium (K⁺) + Hydroxide (OH⁻): KOH

Phosphate: PO₄³⁻ (3 negative charges)

Magnesium (Mg²⁺) + Phosphate (PO₄³⁻): Mg₃(PO₄)₂

Sodium (Na⁺) + Phosphate (PO₄³⁻): Na₃PO₄

Zinc (Zn²⁺) + Phosphate (PO₄³⁻): Zn₃(PO₄)₂

Lithium (Li⁺) + Phosphate (PO₄³⁻): Li₃PO₄

Lead(II) (Pb²⁺) + Phosphate (PO₄³⁻): Pb₃(PO₄)₂

Calcium (Ca²⁺) + Phosphate (PO₄³⁻): Ca₃(PO₄)₂

Iron(II) (Fe²⁺) + Phosphate (PO₄³⁻): Fe₃(PO₄)₂

Iron(III) (Fe³⁺) + Phosphate (PO₄³⁻): FePO₄

Potassium (K⁺) + Phosphate (PO₄³⁻): K₃PO₄

Answer:

• 1 kilocaloric equals 1000 gram calories

• 1 kilocalorie is equals to 4184 Joules of energy.

Explanation:

One kilocalorie is equals to 1000 gram calories because one kilo is equals to or have 1000 grams. One kilocalorie is equals to 4184 Joules of energy while on the other hand, one calorie is equals to 4.184 Joules of energy because one calorie is 1000 times smaller than Kilocalorie so calorie has also 1000 times lower energy than kilocalorie. Kilo is the prefix which means 1000 so we can say that One kilocalorie is equals to 1000 gram calories.