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3 years ago

Two water solutions that have a density of

Chemistry

1 answer:

Colt1911 [192]3 years ago

5 0

<h2>Answer : 627 J</h2><h2>Brainliest</h2>

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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

6 0

3 years ago

Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole

kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

a) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[2 mol]}$

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[1mol]}$

pH = 5,05

I hope it helps!

7 0

3 years ago

Filtration can separate a mixture of fine sand and coarse gravel. The holes in the filter should be large enough to allow ______

Charra [1.4K]

The sand to pass through, but not the gravel. Hope this helps!

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3 years ago

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A rocket is designed to drop its first stage mid-flight. Due to a malfunction, this does not occur. What is a possible result of

artcher [175]

Answer:

The rocket is now too heavy to reach its destination.

3 0

3 years ago

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At what temperature does uranium hexafluoride have a density of 0.9560 g/L at 0.5073 atm?

S_A_V [24]

Given:

Density = 0.7360 g/L.

Pressure = 0.5073 atm.
Step 2
The mathematical expression of an ideal gas is,

Chemistry homework question answer, step 2, image 1
Step 3
Here, R is the universal gas constant (0.0821 L-atm/mol K), T is the temperature in Kelvin, and n is the number of

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4 years ago