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blondinia [14]
3 years ago
13

Determine the volume of a 0.800 M K2Cr2O7 solution required to completely react with 4.24 g of Cu.

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
6 0

<u>Answer:</u> The volume of solution required is 0.0275 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of copper = 4.24 g

Molar mass of copper = 63.55 g/mol

Putting values in above equation, we get:

\text{Moles of copper}=\frac{4.24g}{63.55g/mol}=0.067mol

The chemical equation for the reaction of potassium dichromate and copper follows:

K_2Cr_2O_7+3Cu+7H_2SO_4\rightarrow 3CuSO_4+Cr_2(SO_4)_3+K_2SO_4+7H_2O

By Stoichiometry of the reaction:

3 moles of copper reacts with 1 mole of potassium dichromate.

So, 0.067 moles of copper will react with = \frac{1}{3}\times 0.067=0.022mol of potassium dichromate

To calculate the volume of potassium dichromate, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

We are given:

Moles of potassium dichromate = 0.022 mol

Molarity of solution = 0.800 M

Putting values in above equation, we get:

0.08mol/L=\frac{0.022mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=0.0275L

Hence, the volume of solution required is 0.0275 L.

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What is acide, Base and salt​
ss7ja [257]

Answer:

In acid – base chemistry, salts are ionic compounds that result from the neutralization reaction of an acid and a base. Basic salts contain the conjugate base of a weak acid, so when they dissolve in water, they react with water to yield a solution with pH greater than 7.0.

Explanation:

Hope this helped!

4 0
3 years ago
If you were presented with 2l of a 2m sucrose stock solution, how many grams of sugar would be in a 100 ml aliquot?
Rashid [163]

The 2 L of sucrose stock solution would contain similar concentration with the 100 mL aliquot. Therefore the concentration of aliquot is still 2 M.

The molar mass of sucrose is 342.3 g / mol. Therefore the mass in a 100 mL (0.1 L) aliquot is:

mass = (2 mol / L) * 0.1 L * (342.3 g / mol)

<span>mass = 68.46 g</span>

4 0
3 years ago
A chemistry student is given 250.0 mL of clear aqueous solution at 42 C. He is told an unknown amount of a certain compound X is
lara [203]

Answer:

No

Explanation:

The solubility of a solid in water refers to the amount of that solid that dissolves in water.

It is not possible to calculate the solubility of the solid because the student threw away the first precipitate that formed. We already have the volume of water, but having lost some mass of precipitate, it has become impossible to accurately determine the solubility.

Hence the answer provided above.

7 0
3 years ago
A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2KClO3 mc019-1.jpg 2KCI + 3O2 What is the perc
Anon25 [30]
The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
3 0
3 years ago
Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic ac
Elza [17]

Answer:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1

H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2

HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1

H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2

HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

3 0
3 years ago
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