Which changes when gases condense to form liquids

1 answer:

3 0

In condensation, matter changes from a gas to a liquid. All matter is made of tiny moving particles called molecules. Evaporation and condensation happen when these molecules gain or lose energy. This energy exists in the form of heat.

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You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

8 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$