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AnnyKZ [126]
3 years ago
13

Copper (II) oxide can be heated until it decomposes into elemental copper and oxygen gas. Is 0.695 grams of Copper (II) oxide is

produced from a Copper(II) compound, how many grams of elemental Cu could be produced in the decomposition of CuO?
Chemistry
2 answers:
Oksanka [162]3 years ago
8 0

Answer:

0.555 grams of Cu will be produced

Explanation:

Step 1: Data given

Mass of copper (II) oxide = 0.695 grams

Molar mass Copper (II) oxide = 79.545 g/mol

Step 2: The balanced equation

2CuO → 2Cu + O2

Step 3: Calculate moles CuO

Moles CuO = mass CuO / molar mass CuO

Moles CuO = 0.695 grams / 79.545 g/mol

Moles CuO = 0.00874 moles

Step 4: Calculate moles Cu

For 2 moles CuO we'll have 2 moles Cu and 1 mol O2

For 0.00874 moles CuO we'll have 0.00874 moles

Step 5: Calculate mass Cu

Mass Cu = 0.00874 moles * 63.546 g/mol

Mass Cu = 0.555 grams

0.555 grams of Cu will be produced

insens350 [35]3 years ago
4 0

Answer:

0.55 g of Cu can produced by this decomposition

Explanation:

Let's think the reaction of decomposition:

2CuO → 2Cu + O₂

Ratio is 2:2. 2 moles of CuO can decompose into 2 moles of Cu and 1 mol of oxygen.

Let's convert the mass of oxyde to moles → 0.695 g. 1 mol/ 79.55 g =

8.74×10⁻³ moles

8.74×10⁻³ mol of CuO decompose to 8.74×10⁻³ moles of Cu

Let's find out the mass → 8.74×10⁻³ mol . 63.55 g/1mol = 0.55 g

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2 years ago
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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
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Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

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Answer:

A. 6N

B. 4H, 2O

C. 4H, 4N, 12O

D. 2Ca, 4O, 4H

E. 3Ba, 6Cl, 18O

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6 0
3 years ago
At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar
Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

 

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

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Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

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Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

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P(acetic acid) = 73.45 torr ≈73.5

 

We can control this by calculating the partial pressure of methanol

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226 - 152.55 = 73.45 torr  

The partial pressure of acetic acid is 73.5 torr

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