Aqueous solutions of copper (II) bromide and silver (1) acetate react to form solid

Chemistry

1 answer:

Sladkaya [172]4 years ago

4 0

Answer:

5.63 g

Explanation:

Step 1: Write the balanced equation

CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)

Step 2: Calculate the reacting moles of copper (II) bromide

30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:

$0.0300L \times \frac{0.499mol}{L} = 0.0150mol$

Step 3: Calculate the moles formed of silver (I) bromide

The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.

Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr

The molar mass of AgBr is 187.77 g/mol.

$0.0300 mol \times \frac{187.77g}{mol} = 5.63 g$

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At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate

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Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$ .

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Therefore, $K_{c''}$ = $K_{c'}$ = $(20)^{0.5}$

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As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.