All categories

2 years ago

Formula of preparation of sodium hydroxide

1 answer:

7 0

The chemical formula of sodium hydroxide is NaOH, and its molar mass is 40.01 g/mol. It is the alkali salt of sodium, and its structure is shown below:

It is an ionic compound consisting of sodium cation (Na+) and hydroxide (OH-) anion.

You might be interested in

What is the chemical test to differentiate ethanol from ethanoic acid

Tomtit [17]

Answer:

Sodium hydrogencarbonate or sodium carbonate

8 0

2 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago

10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.

ehidna [41]

<u>Answer: </u>The molar mass of solute is 115 g/mol.

<u>Explanation:</u>

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

The expression for the calculation of elevation in boiling point is:

$\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m$

$\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Boiling point of pure solvent (benzene) = $80.10^oC$

Boiling point of solution = $81.20^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_b$ = Boiling point elevation constant = $2.53^oC/m$

$m_{solute}$ = Given mass of solute = 10 g

$M_{solute}$ = Molar mass of solute = ? g/mol

$w_{solvent}$ = Mass of solvent = 200 g

Putting values in equation 1, we get:

$81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol$

Hence, the molar mass of solute is 115 g/mol.

4 0

3 years ago

Write a essay on endangered animals

Kay [80]

Answer:

WHATS OOP Lucy???

7 0

2 years ago

Will a precipitate form if you mix barium chloride and potassium carbonate?

kaheart [24]

Potassium carbonate reacts with barium chloride to precipitate insoluble barium salts that is barium carbonate, when the solutions are mixed together you'll see white precipitate that is barium carbonate and potassium is dissolve in the solution to form potassium chloride.

5 0

2 years ago

Formula of preparation of sodium hydroxide​

Formula of preparation of sodium hydroxide