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3 years ago

Calculate the formula unit mass of CaCl2

1 answer:

3 0

Formula unit mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the empirical formula of a compound. It is expressed in amu.

Atomic mass of calcium = 40 amu

Atomic mass of chlorine = 35.5 amu

Formula mass of CaCl2 = (1 x 40) + (2 x 35.5) = 111amu.

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4. How does the water vapor change as altitude increases?

Karolina [17]

Answer:

it drifts through the atmosphere and the vapor absorbs heat then the warm air rises up therefore leaving the planet's surface and releasing the heat back into the upper part of the atmosphere

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]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

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

Essential amino acids cannot be made

sattari [20]

Answer:

True

Explanation:

There are 9 essential amino acids which cannot be made by the body and have to be obtained from food.

They are histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine

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3 years ago

PLEASE HELP!

Eduardwww [97]

Answer:

The answer is C, number 4 shows a rift valley

Explanation: This is because the shape of the feature is literally a rift and it forms a valley lol. Also i have some evidence:

7 0

3 years ago

Read 2 more answers

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Morgarella [4.7K]

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5% as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

$\% H=\frac{m_H*1}{M_{HF}}*100\%$

Where $m_H$ stands for the atomic mass of hydrogen and $M_{HF}$ for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

$\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%$

Learn more about percent compositions: brainly.com/question/12247957

5 0

3 years ago

Calculate the formula unit mass of CaCl2​

Calculate the formula unit mass of CaCl2