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8090 [49]
3 years ago
9

Calculate the mass of 0.0455 mol of Ni

Chemistry
1 answer:
Lera25 [3.4K]3 years ago
8 0

Answer:

2.67g Ni

Explanation:

To convert from moles to mass, use molar mass as a conversion factor.

Nickel has a molar mass of 58.69g/mol.

0.0455mol (\frac{58.69g Ni}{1 mole Ni}) = 2.67g Ni

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3 years ago
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}.

Rewrite as a quadratic equation and solve for x:

x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)

x\approx 0.00111.

The pH of a solution depends on its H⁺ concentration.

At equilibrium

[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}.

\text{pH} = -\log{[\text{H}^{+}]} = 2.79.

5 0
3 years ago
A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?
Sindrei [870]

A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O

Let the total mass of compound = 100g

The mass of Aluminum = 22.10 g

Moles of Al = mass / atomic mass = 22.10 /23= 0.96

The mass of P = 25.40 g

Moles of P = 25.40 / 31 = 0.819

The mass of oxygen  = 52.50 g

Moles of oxygen = 52.50 / 16= 3.28

The mole ratio of the two elements will be

Al = 0.96/0.819 = 1.17

P = 0.819/0.819 = 1

O = 3.28 / 0.819 = 4

The whole number ratio will be 1 : 1 : 4

So formula will be AlPO4


5 0
3 years ago
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