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4 years ago

6

How does adding a non-volatile solute to a pure solvent affect the boiling point of the pure solvent?

1 answer:

belka [17]4 years ago

7 0

The correct answer is the second statement. The solvent will have a higher boiling point. Adding a non-volatile solute to a pure solvent will increase the boiling point of the solvent. This solution exhibit colligative properties. Colligative properties depend on the amount of solute dissolved in a solvent. These set of properties do not depend on the type of species present.

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sashaice [31]

D. I, II, and IV. Basically everything but the temperature.

6 0

4 years ago

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 8.60 MJ o

Alecsey [184]

Answer: $64.216\times 10^4grams$ of liquid sodium is required.

Explanation: To calculate the mass of liquid sodium, we use the formula:

$q=nc\Delta T\\q=\frac{m}{M}c\Delta T$

where,

q = heat required, $8.60MJ=8.60\times 10^6J$ (Conversion factor: 1MJ = 1000000J)

m = Mass of liquid sodium, $?g$

M = Molar mass of liquid sodium, $23g/mol$

c = Specific heat capacity, $30.8J/Kmol$

$\Delta T$ = change in temperature, $\Delta T=(510-500)K=10K$ (Conversion factor: 0°C = 273K)

Putting values in above equation, we get:

$8.60\times 10^6J=\frac{m}{23g/mol}\times 30.8J/Kmol\times 10K$

$m=64.216\times 10^4grams$

5 0

3 years ago

A student prepared a stock solution by dissolving 20.0 g of NaOH in enough water to make 150. mL of solution. She then took 15.0

pav-90 [236]

Answer:

$0.769\ \text{M}$

Explanation:

Mass of stock solution = 20 g

Molar mass of NaOH = 40 g/mol

Volume of stock solution = 0.150 mL

$M_2$ = Concentration of NaOH for the final solution

$V_1$ = Amount of stock solution taken = 15 mL

$V_2$ = Total volume of solution = 65 mL

Molarity is given by

$M_1=\dfrac{\text{Mass}}{\text{Molar mass}\times \text{Volume}}\\\Rightarrow M_1=\dfrac{20}{40\times 0.15}\\\Rightarrow M_1=\dfrac{10}{3}$

We have the relation

$M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{\dfrac{10}{3}\times 15}{65}\\\Rightarrow M_2=0.769\ \text{M}$

The concentration of NaOH for the final solution is $0.769\ \text{M}$ .

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3 years ago

A student measures the frequency of a wave as 2 waves/second, and the wavelength as 40 cm/wave. Use the velocity equation (v = f

Dafna1 [17]

V = 2*40
V= 80 m/s
I am not sure!!!

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3 years ago

Please Help In need of chemistry help

Firlakuza [10]

It’s an example of substitution

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3 years ago