Which of the following contribute to acid rain

Chemistry

1 answer:

Otrada [13]3 years ago

6 0

I don't see any options can you add them?

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yulyashka [42]

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2 years ago

As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures lis

Setler79 [48]

Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2

when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.

c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.

3 0

3 years ago

The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.36 ml of water at this temperature?

spin [16.1K]

Mass of 1 ml of water 1 g as density of water 1.00 g/ml at 4° C. So, mass of 2.36 ml of water $\frac{1}{2.36}$ g= 0.423 g.

Molecular mass of water is 18 g which indicates 18 g water contains 6.023 X $10^{23}$ number of water molecules. So, 0.423 g of water contains $\frac{6.023 X 10^{23} X 0.423 }{18} g$ = 0.141 X $10^{23}$ number of water molecules= 1.41 X $10^{22}$ number of water molecules.