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3 years ago

Place the following substances in order of increasing vapor pressure at a given temperature.

Chemistry

1 answer:

Butoxors [25]3 years ago

8 0

Answer:

NH3 < NF3 < BCl3

Explanation:

The vapour pressure of a substance has something to do with the nature of intermolecular forces between its molecules. If the molecules of a substance are held together by strong intermolecular forces, the substance will display a low vapour pressure at a given temperature and vice versa.

Ammonia has the lowest vapour pressure because of strong intermolecular hydrogen bonds that hold its molecules together.

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17.8g of aluminum sulfate is produced from how many grams of potassium sulfate?

german

Answer:

1 mole of ferric contains 2 moles of iron,and 12 moles of oxygen atoms, and three moles of sulphate ions

3 0

3 years ago

A sample of nitrogen occupies 10.0 liters at 25°C and 98.7 kPa. What would be the volume at 20°C and 102.7 kPa?

krek1111 [17]

To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.

n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2

At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:

V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L

7 0

3 years ago

Why do Hurricanes not form directly on the equator?

Gemiola [76]

They do form directly on the equators

8 0

3 years ago

A liquid has an empirical formula CCl2, and a boiling point of 1 21 oC. When vapourised, the gaseous compound has a density of 4

Darya [45]

Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

Pressure = 785 torr = 1.033 atm

density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

Learn more about molar mass of a gas at: brainly.com/question/26215522

6 0

1 year ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago