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AVprozaik [17]
3 years ago
10

Calculate the molar concentration of the Cl⁻ ions in 0.73 M MgCl2(aq).

Chemistry
1 answer:
Marta_Voda [28]3 years ago
7 0
MgCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MgCl2 that dissolves.
MgCl2(s) --> Mg+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.73 mol MgCl2/1L × 2 mol Cl⁻ / 1 mol MgCl2 = 1.5 M
The answer to this question is [Cl⁻] = 1.5 M
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chegg 2. What pattern did you observe measuring cell voltages with a silver electrode versus with a platinum/H2 electrode
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7 0
2 years ago
The reaction H2(g)+F2(g)?2HF(g) is spontaneous at all temperatures. A classmate calculates the entropy change for this reaction
Pachacha [2.7K]

Answer:

The correct answer is there is a mistake in the calculation. The second law of thermodynamics state that in any spontaneous process there is an increase in the entropy of the universe.

Explanation:

According to the second law a reaction will occur in a system spontaneously if the total entropy of both system and surrounding increases during the reaction.That means in case of spontaneous reaction entropy change is always positive.

     But according to the question the reaction H2+F2=2HF is spontaneous in all temperature.So according to the second law of thermodynamics i can say that my classmate made a mistake in calculation that"s why his result for entropy change comes negative.

6 0
3 years ago
A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

HCl+NaOH\rightarrow NaCl+H_2O

For neutralization:

1 mole of H^+ ion will react with 1 mole of OH^- ion

0.01 mol of H^+  ion will react with =\frac{1}{1}\times 0.01mole of OH^- ion

Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

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pH+pOH=14

pH=14-1=13

Thus the pH of resulting solution will be 13

7 0
3 years ago
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