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svet-max [94.6K]
3 years ago
7

Match the following aqueous solutions with the appropriate letter.

Chemistry
1 answer:
Alla [95]3 years ago
6 0

Answer:

Explanation:

When a salt is dissolved , it increases  the boiling point . Increase in boiling point depends upon number of ions . So it is a colligative property .

.19 m AgNO₃ . Each molecule  will ionize into two ions . So effective molar concentration is 0.19 x 2 = .38 m

0.17 m CrSO4.Each molecule  will ionize into two ions . So effective molar concentration is 0.17 x 2 = .34 m

0.13 m Mn(NO₃)₂. Each molecule  will ionize into three  ions . So effective molar concentration is 0.13 x 3 = .39 m

0.31 m Sucrose(nonelectrolyte). Molecules will not ionize . So effective molar concentration is 0.31 x 1 = .31 m

Higher the molar concentration , greater the depression in boiling point .

So lowest boiling point is 0.13 m Mn(NO₃)₂.

second highest  boiling point is 0.19 m AgNO3.

Third lowest boiling point is 0.17 m CrSO4

Highest boiling point or lowest depression 0.31 m Sucrose.

a . 4

b .  1

c . 2

d . 3

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Answer:

6.15.3 k

Explanation:

From the question we can see that

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putting values wet

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Guanidin HNC(NH2)2 is a fertilizer. What is the percent by mass of nitrogen in the fertilizer
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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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Answer: The concentration of hydrogen ions for this solution is 1.99 \times 10^{-3}.

Explanation:

Given: pOH = 11.30

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Also, pH is the negative logarithm of concentration of hydrogen ions.

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pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}

Thus, we can conclude that the concentration of hydrogen ions for this solution is 1.99 \times 10^{-3}.

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