All categories

4 years ago

54 points

Chemistry

2 answers:

xeze [42]4 years ago

7 0

Answer: The correct answer is temperature.

Explanation:

The solubility is defined as the amount of solute that can be dissolved in a given amount of solvent at specific temperature.

We are given that the solubility of NaCl is 35.9g per 100g at 20°C.

If we dissolve 35.9 grams in 100 grams of water, it is completely dissolve at this temperature.

If the temperature increases, the solubility also increases and if temperature decreases, the solubility decreases.

It is given that some solid remains undissolved, so the temperature would have been decreased, therefore the solubility decreases.

Hence, the correct answer is temperature.

Andreyy894 years ago

5 0

the correct answer is temperature

You might be interested in

Which part of making a solution always releases energy?

emmainna [20.7K]

The part of making a solution that always releases energy is the overall change in forming the solution. The answer is letter D. Although letters A, B and C can be viable answers but, it is not always the case. There are some substances that when you mix or separate them requires more energy or less energy. An example would be when the formation (or enthalpy of formation) of carbon dioxide is negative, it means that it releases heat to the surroundings. When it releases heat to the surroundings, the reaction is exothermic. Another example is when you mix baking soda and muriatic acid, the resulting mixture is colder. When it is cold, it means that the reaction is endothermic. So the best answer is letter D.

4 0

3 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

Answer: The freezing point of solution is -0.974°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

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

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

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

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

Determine the number of equivalents if a 3.89N solution contains 0.76 L of solution

ValentinkaMS [17]

Answer:

Explanation:

oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.

5 0

3 years ago

PLEASE I NEED YOUR HELP, I WILL GIVE 10 POINTS PLEASEEEEE

Cerrena [4.2K]

Answer:

Erosion

Explanation:

3 0

3 years ago

Analyze the following blood splatter pictures to determine the general direction, origin, and impact

kkurt [141]

Answer:

!!!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago