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3 years ago

When 1.187 g of a metallic oxide is reduced with excess hydrogen 1.054 g of the metal is produced. what is the metallic oxide?

Chemistry

2 answers:

Ira Lisetskai [31]3 years ago

8 0

Answer: The metallic oxide formed will be $Cu_2O$

Explanation:

We are given a metallic oxide, having general chemical formula $M_2O_n$

We are given:

Mass of metallic oxide = 1.187 g

Mass of metal = 1.054 g

Mass of oxygen = 1.187 - 1.054 = 0.133 g

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .......(1)

Given mass of oxygen = 0.133 g

Molar mass of oxygen = 16 g/mol

Putting values equation 1, we get:

$\text{Moles of oxygen}=\frac{0.133g}{16g/mol}=0.0083mol$

Number of moles of metal in the oxide is twice than the number of moles of oxygen

Number of moles of metal = $(2\times 0.0083)=0.0166$ moles

Now, calculating the molar mass of metal by using equation 1, we get:

Moles of metal = 0.0166 moles

Mass of metal = 1.054 g

Putting values in equation 1, we get:

$0.0166mol=\frac{1.054g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.054g}{0.0166mol}=63.49g/mol$

The metal having 63.49 g/mol as molar mass is copper

Hence, the metallic oxide formed will be $Cu_2O$

Anit [1.1K]3 years ago

6 0

Mass O = 1.187 - 1.054 = 0.133 g
Moles O = 0.133 g / 16 g/mol = 0.00831 
Moles metal = 2 x 0.00831 = 0.0166 
Atomic mass metal = 1.054 g / 0.0166 = 63.4 g/mol 
( this is the atomic mass of Cu) 
so the metal oxide is Cu2O

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .