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3 years ago

How many moles of compound are there in the following?

Chemistry

1 answer:

slavikrds [6]3 years ago

8 0

a) (NH4)2SO4 --- 1 mole of it contains 2 moles of N, 8 moles of H, 1 mole of S, and 4 moles of O.

MM = (2 moles N x 14.0 g/mole) + (8 moles H x 1.01 g/mole) + (1 mole S x 32.1 g/mole) + (4 moles O x 16.0 g/mole) = 132 g/mole.

6.60 g (NH4)2SO4 x (1 mole (NH4)2SO4 / 132 g (NH4)2SO4) = 0.0500 moles (NH4)2SO4

b) The molar mass for Ca(OH)2 = 74.0 g/mole, calculated like (NH4)2SO4 above.

4.5 kg Ca(OH)2 x (1000 g / 1 kg) x (1 mole Ca(OH)2 / 74.0 g Ca(OH)2) = 60.8 moles Ca(OH)2

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The rate constant for a certain reaction is k = 6.50×10−3 s−1 . If the initial reactant concentration was 0.600 M, what will the

Kamila [148]

Answer: 1.037M

Explanation:

Since the rate constant unit is per seconds, therefore it is a first order reaction.

First order reaction equation is given as

InA= -kt +InAo

Where,Ao is the initial concentration of reactant =0.600M

A is the concentration of reactant at a specifies time t=3×60=180s

and k is the rate constant

InA = -6.50×10^-3 ×180 +In(0.6)

InA = -1.17 + 0.5108

InA= -1.680

A = e-1.680

A= 1.037M

Therefore the concentration after 3minutes is 1.037M

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3 years ago

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Explanation:

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3 years ago

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