How do the scientist know that cells exist

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

Formula Silver(II) Oxide

Svetach [21]

Answer:

AgO

Explanation:

5 0

3 years ago

In the sn2 experiment, what was the purpose of washing the distilled product with 5% naoh

photoshop1234 [79]

<span>The purpose of washing the product with NaOH is simply to neutralize any acid which remained or leaked after the 1st initial separation. The NaOH base reacts with the acid to form neutralization reaction products which are soluble in water.</span>

5 0

3 years ago

What is the idea behid boyles law

zavuch27 [327]

Answer:

Explanation:

For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. Or Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship. If volume increases, then pressure decreases and vice versa, when the temperature is held constant.

3 0

2 years ago

For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--> N2(g) + 3H2(g) K, 0.83 1. When

erma4kov [3.2K]

Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where K is defined as:

$K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83$

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

$Q = \frac{1atm*{1atm}^3}{1atm^2} = 1$

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

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4 0

3 years ago