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Alenkasestr [34]
2 years ago
9

how does activity explain the differences in the observed solubility of Ca(IO3)2 into each of these solvents

Chemistry
1 answer:
avanturin [10]2 years ago
3 0

There are differences in the observed solubility of calcium iodate (Ca(IO3)2) in different types of solvent due to its unique composition.

There differences in the observed solubility of Ca(IO3)2 into each of these solvents because calcium iodate (Ca(IO3)2) is not soluble in all types of solvent. In some solvents, it is fully soluble whereas in other solvents it is insoluble or slightly soluble.

For example, calcium iodate (Ca(IO3)2) is soluble in nitric acid but insoluble in alcohol due to its unique structure and composition so we can conclude that there are differences in the observed solubility of calcium iodate (Ca(IO3)2) in different types of solvent due to its unique composition.

Learn more: brainly.com/question/25798049

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For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
When may a scientific theory be revised?
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Answer:

it can be revised at any time.

Explanation:

6 0
4 years ago
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Explain what you think a compound is
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Answer:

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Explanation:

6 0
3 years ago
7. Suppose 1.01 g of iron (III) chloride is placed in a 10.00-mL volumetric flask with a bit of water in it. The flask is shaken
Nana76 [90]

<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.

<u>Explanation:</u>

Molarity is defined as the number of moles present in one liter of solution.  The equation used to calculate molarity of the solution is:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Or,

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M

Hence, the molarity of Iron (III) chloride is 0.622 M.

3 0
3 years ago
A 1.250-g sample of benzoic acid, C7H6O2, was placed in a combustion bomb. The bomb was filled with an excess of oxygen at high
Degger [83]

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>

1.250g * (1mol / 122.12g) = 0.0102 moles

<em />

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

<em>3224 kJ/mol </em>

4 0
3 years ago
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