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2 years ago

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If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting so

lution?
A. 0.20 M Mg ion & 0.20 M Cl ionB. 0.40 M Mg ion & 0.40 M Cl ionC. 0.20 M Mg ion & 0.40 M Cl ionD. 2.0 M Mg ion & 2.5 M Cl ion

1 answer:

sladkih [1.3K]2 years ago

8 0

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂ → Mg²⁺ + 2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

MgCl₂ → Mg²⁺ + 2Cl⁻

0.12mol 0.12 0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

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Explanation:

we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹

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The average heat evolved by the oxidation of foodstuffs is then:

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⇒ ΔT = 5.16 K

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A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

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2 years ago

What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi

kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

$\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s)$ .

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

$\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq)$ .

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

How to rewrite a chemical equation to produce a net ionic equation?

Rewrite all reactants and products that ionizes completely in the solution as ions.
Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

Soluble salts,
Strong acids, and
Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as $\rm Mg^{2+}$ and $\rm 2\; Cl^{-}$ .
NaOH is a strong base. Each unit of NaOH can be written as $\rm Na^{+}$ and $\rm OH^{-}$ .
Mg(OH)₂ is a weak base and should not be written.
NaCl is a salt and is soluble in water. Each unit of NaCl can be written as $\rm Na^{+}$ and $\rm Cl^{-}$ .

$\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}$ .

Ions on both sides of the equation:

$\rm 2\; Cl^{-}$ , and
$\rm 2\; Na^{+}$ .

Add the state symbols:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

For the second reaction:

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

$\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}$ .

$\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s)$ .

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2 years ago

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