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Inessa [10]
3 years ago
5

Potassium carbonate, K2CO3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH3OH, and ammonium chloride, NH4Cl, are solub

le in water. Which produces the largest number of dissolved particles per mole of dissolved solute?
A) K2CO3B) NaI
C) MgCl2
D) CH3OH
E) NH4Cl
Chemistry
1 answer:
just olya [345]3 years ago
5 0

Answer:

A. K₂CO₃

Explanation:

The chemical reaction showing the ions produced on dissolving Potassium carbonate in water is:

K₂CO₃ ⇒ 2K⁺ + CO₃²⁻

Number of ions produced = 2 + 1 = 3

The chemical reaction showing the ions produced on dissolving sodium iodide in water is:

NaI ⇒ Na⁺ + I⁻

Number of ions produced = 1 + 1 = 2

The chemical reaction showing the ions produced on dissolving potassium bromide in water is:

KBr ⇒ K⁺ + Br⁻

Number of ions produced = 1 + 1 = 2

The chemical reaction showing the ions produced on dissolving ammonium chloride in water is:

NH₄Cl ⇒ NH₄⁺ + Cl⁻

Number of ions produced = 1 + 1 = 2

<u>The largest number of dissolved particles per mole of dissolved solute is produced by Potassium carbonate.</u>

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The process of spoilage is speeded up by bananas for example, giving up Ethylene gas. You do not want to put a banana with tomatoes, because tomatoes are very sensitive to Ethylene. (It's OK to eat them together. They make a terrific salad. Yum).

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3 0
3 years ago
Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid wate
allochka39001 [22]

Answer:

a) 2NaOH+H2SO4→Na2SO4+2H2O2NaOH+H2SO4→Na2SO4+2H2O

b) Số phân tử NaOH : Số phân tử H2SO4 = 2:1

Số phân tử NaOH : Số phân tử Na2SO4 = 2:1

Số phân tử NaOH : Số phân tử H2O = 2:2

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8 0
2 years ago
What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?
8090 [49]
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
3 0
3 years ago
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