Which substance contains a bond with the greatest ionic character?

When heated to 150 ºC, CuSO4.5 H2O loses its water of hydration as gaseous H2O. A 2.50 g sample of the compound is placed in a s

Katyanochek1 [597]

Answer:

Water pressure 0.5 atm

Total Pressure= 2.27 atm

Explanation:

To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.

First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:

P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁

For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:

PV = nRT ∴ P = nRT/V

n will we calculated from the quantity of sample.

2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O

the amount water of hydration is

= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O

= 0.05 mo H₂O

pressure of dry air at the final temperature,

P₂ = 1 atm x 500 K/ 300 K = 1.67 atm

Pressure of water :

P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm

∴ Total Pressure = 1.67 atm

H2O Pressure = 0.5 atm

5 0

3 years ago

Which of the following is true about the distribution of water in the hydrosphere?

mr Goodwill [35]

The answer is: C. The amount of water in the hydrosphere remains constant.

The hydrosphere denotes the entire water available worldwide in lakes and streams, subsurface water, the ice of glaciers, and water vapor in the atmosphere. The water cycle is the process that converts liquid water to atmospheric water vapor (and vice versa) throughout the world.

5 0

3 years ago

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b

Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

Step 1: Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

Step 2: Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

Step 3: Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

Step 4: Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

5 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

Do elements in the same group have same principle number?

Leviafan [203]

The s-, p-, and d-block elements of the periodic table are arranged into 18 numbered columns, or groups. The elements in each group have the same number of valence electrons. As a result, elements in the same group often display similar properties and reactivity.

4 0

3 years ago

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