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3 years ago

(Thermodynamics)

Chemistry

1 answer:

frutty [35]3 years ago

3 0

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

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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:

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

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.

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

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3 years ago

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sleet_krkn [62]

Answer:

V1= 0.305L

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1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

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3 years ago

How many particles are in 3.2 mole of neon gas

Cerrena [4.2K]

There are 1.93 x 10²⁴ particles

<h3>Further explanation</h3>

Given

3.2 moles of Neon gas

Required

Number of particles

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance

<em>1 mol = 6.02.10²³ particles </em>

Can be formulated

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