Balance each chemical equation. Hg2(C2H3O2)2(aq)+KCl(aq)→Hg2Cl2(s)+KC2H3O2(aq)

Chemistry

1 answer:

Ira Lisetskai [31]2 years ago

4 0

Answer:

Explanation:

When potassium chloride reacts with lead acetate in aqueous medium, it gives white precipitate of lead chloride and potassium acetate in aqueous medium as a product.

As shown in chemical reaction given below:

You might be interested in

What is diffusion?<br>Class 9th <br>Chemistry <br><br><br><br><br>

tester [92]

Diffusion is the movement of a substance from an area of high concentration to an area of low concentration . Diffusion happens in liquids and gases because their particles move randomly from place to place . Diffusion is an important process for living things ; it is how substances move in and out of cells .

Pls follow me and Mark as brainlest :-)

6 0

3 years ago

Read 2 more answers

Write the fraction of the mass of kcl produced from 1 g of k2c03

Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$