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OLEGan [10]
2 years ago
14

Identify the reactants and products and write chemical equations for each of the following reactions: a. Gaseous chlorine reacts

with an aqueous solution of potassium bromide to form liquid bromine and an aqueous solution of potassium chloride. b. Solid aluminum reacts with solid iodine to produce solid aluminum iodide. c. Solid magnesium reacts with an aqueous solution of hydrochloric acid to form an aqueous solution of magnesium chloride and bubbles of hydrogen gas.
Chemistry
1 answer:
Aneli [31]2 years ago
6 0

Answer:

a. Cl (g) + KBr (aq) = Br (l) + KCl (aq)

b. Al (s) + I (s) = AlI (s)

c. Mg (s) + HCl (aq) = MgCl (aq) + H2 (g)

Explanation:

The reactants are the elements before the equals sign and the products are the ones after the equals sign that are bolded.

Hope this helps :)

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Llana [10]

Answer:

V_{HCl}=0.208L=208mL

Explanation:

Hello,

In this case, since the chemical reaction is:

2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

n_{HCl}=2*n_{Mg(OH)_2}

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g}  =0.00858mol

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M

Then, since the concentration and the volume define the moles, we can write:

[HCl]*V_{HCl}=2*n_{Mg(OH)_2}

Therefore, the neutralized volume turns out:

V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL

Best regards.

3 0
3 years ago
Cu2 -> Cu2+ is it getting oxidized?
KIM [24]

Answer:

Explanation:

Cu²⁺ + 2e⁻  →  Cu   ( copper gets reduced )

Cu  → Cu²⁺ + 2e⁻ (  copper gets oxidized )

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Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

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Answer:

physical

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