BOLD ARE ANSWER CHOICES. THANKS. #3. At 667K, HI is found to be 11.4% dissociate

Question

BOLD ARE ANSWER CHOICES. THANKS.

#3. At 667K, HI is found to be 11.4% dissociated into its elements.

2 HI (g) ßà H2 (g) + I2 (g)

       If 1.00 mol HI is places in a 1.00 L container and heated to 667K, calculate the

       value of Kc for this equilibrium at this temperature.

3.67x10–3

4.14x10–3

1.30x10–2

68.2

76.9

#4. Consider the equilibrium

N2 (g) + O2 (g) ßà 2 NO (g)

       At 2300 K the equilibrium constant Kc = 1.70x10–3. If 0.15 mol NO (g) is placed

       into a 10.0 L flask and heated to 2300K, what is the [NO]?

3.03x10–4 M

3.82x10–2 M

7.35x10–2 M

7.65x10–2 M

0.147 M

#6. Determine the [OH–] and pH when 15.0 g of Ba(OH)2 is dissolved in water to

        produce 250. mL of solution.

[OH–] = 0.350 M; pH = 13.54

[OH–] = 0.350 M; pH = 0.46

[OH–] = 0.700 M; pH = 0.16

[OH–] = 0.700 M; pH = 13.84

[OH–] = 0.175 M; pH = 13.24

9. To buffer a solution at a pH = 4.57, what mass of sodium acetate (NaCH3COO)

       should you add to 500. mL of a 0.150 M solution of acetic acid (CH3COOH).

       Ka = 1.75x10–5.

2.86 g

3.97 g

6.86 g

8.61 g

9.53 g

Explanation / Answer

The equation is : 2 HI (g) <-----> H2 (g) + I2 (g)

1.00 mol of HI is placed in 1.00 L container. So initial [HI] = 1.00 M

Dissociation is 11.4 %.

So 11.4 % of 1.00 M = (11.4/100) * 1.00 = 0.114 M

2 mol of HI decomposes giving 1 mole of H2 and 1 mole of I2

So 0.114 M of HI decomposes giving 0.114/2 = 0.057 M of H2 and I2

remaining [HI] = 1.00 - 0.114 = 0.886 M

Kc = ( [H2] [I2])/ [HI]^2

= [ (0.057*0.057)] / (0.886)^2

= 4.41*10-3

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