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Pepsi [2]
3 years ago
8

Write the equation for the formation of a fluoride anion from a fluorine atom. If the electron affinity of the fluorine atom is

328 kJ/mol, what is q when 2.5 mol of fluoride anions are formed from fluorine atoms? How much energy is released when one F atom accepts an electron?
Chemistry
1 answer:
Finger [1]3 years ago
5 0

Answer:

Part A. F + e⁻ → F⁻

Part B. -820 kJ

Part C. -5.45x10⁻²² kJ

Explanation:

Part A

The fluoride anion is formed when fluorine accepts an electron:

F + e⁻ → F⁻

Part B

The electron affinity is the energy necessary for the atom to accept 1 electron, which is exothermic, so it's negative. The total energy q is the electron affinity multiplied by the number of moles:

q = -328 kJ/mol * 2.5 mol

q = -820 kJ

Part C

By the Avogadros' number

1 mol --------- 6.02x10²³ atoms

x --------- 1 atom

By a simple direct three rule:

x = 1.66x10⁻²⁴ mol

So, the energy for 1 atom is:

q = -328 kJ/mol * 1.66x10⁻²⁴ mol

q = -5.45x10⁻²² kJ

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If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 127oC in an evacuated
zhuklara [117]

Answer : The the partial pressure of the nitrogen gas is 0.981 atm.

The total pressure in the tank is 2.94 atm.

Explanation :

The balanced chemical reaction will be:

(CH_3)_2N_2H_2(l)+2N_4O_4(l)\rightarrow 3N_2(g)+4H_2O(g)+2CO_2(g)

First we have to calculate the moles of dimethylhydrazine.

Mass of dimethylhydrazine = 150 g

Molar mass of dimethylhydrazine =60.104 g/mole

\text{Moles of dimethylhydrazine}=\frac{\text{Mass of dimethylhydrazine}}{\text{Molar mass of dimethylhydrazine}}

\text{Moles of dimethylhydrazine}=\frac{150g}{60.104g/mole}=2.49mole

Now we have to calculate the moles of N_2 gas.

From the balanced chemical reaction we conclude that,

As, 1 mole of (CH_3)_2N_2H_2 react to give 3 moles of N_2 gas

So, 2.49 mole of (CH_3)_2N_2H_2 react to give 2.49\times 3=7.47 moles of N_2 gas

Now we have to calculate the partial pressure of nitrogen gas.

Using ideal gas equation :

PV=nRT\\\\P_{N_2}=\frac{nRT}{V}

where,

P = Pressure of N_2 gas = ?

V = Volume of N_2 gas = 250 L

n = number of moles  N_2 gas = 7.47 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 127^oC=273+127=400K

Putting values in above equation, we get:

P_{N_2}=\frac{(7.47mole)\times (0.0821L.atm/mol.K)\times 400K}{250L}=0.981atm

Thus, the partial pressure of the nitrogen gas is 0.981 atm.

Now we have to calculate the total pressure in the tank.

Formula used :

P_{N_2}=X_{N_2}\times P_T

P_T=\frac{1}{X_{N_2}}\times P_{N_2}

P_T=\frac{1}{(\frac{n_{N_2}}{n_T})}\times P_{N_2}

P_T=\frac{n_{T}}{n_{N_2}}\times P_{N_2}

where,

P_T = total pressure = ?

P_{N_2} = partial pressure of nitrogen gas = 0.981 atm

n_{N_2} = moles of nitrogen gas = 3 mole  (from the reaction)

n_{T} = total moles of gas = (3+4+2) = 9 mole  (from the reaction)

Now put all the given values in the above formula, we get:

P_T=\frac{9mole}{3mole}\times 0.981atm=2.94atm

Thus, the total pressure in the tank is 2.94 atm.

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Help me with this questions please
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I will figure out the answer and i will letu know what else do u want to talk about .
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