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3 years ago

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What is the predominant intermolecular force in the liquid state of each of these compounds: ammonia (NH3), methane (CH4), and n

itrogen trifluoride (NF3)?

1 answer:

MAVERICK [17]3 years ago

3 0

Answer:

The predominant intermolecular force in the liquid state of each of these compounds:

ammonia (NH3)

methane (CH4)

and nitrogen trifluoride (NF3)

Explanation:

The types of intermolecular forces:

1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.

2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.

3. London dispersion forces exist between all the atoms and molecules.

NH3 ammonia consists of intermolecular H-bonding.

Methane has London dispersion forces.

Because both carbon and hydrogen has almost similar electronegativity values.

NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

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3 years ago

Consider the reaction of Mg3N2 with H2O to form Mg(OH)2 and NH3. If 4.33 g H2O is reacted with excess Mg3N2 and 6.26 g of Mg(OH)

Len [333]

Answer:

89.34%

Explanation:

First, write a balanced reaction.

Mg3N2 + <u>6</u>H2O --> <u>3</u>Mg (OH)2 + <u>2</u>NH3

Next determine the moles of the known substance, or limiting reagent ( H2O)

n= m/MM

n ( H2O) = 4.33/(1.008×2)+16

n(H2O)= 0.2403

Use the mole ratio to find the moles of Mg(OH)2

0.2403 ÷2

n (Mg (OH)2) = 0.1202

Next, find the theoretical mass of Mg (OH)2 that should have been produced

m= n × MM

m= 0.1202 × (24.305 + (16×2) +(1.008 ×2))

=7.007g

To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.

6.26/ 7.007 × 100

=89.34%

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