Question

What did I do wrong?

1) A 44.3 g piece of Aluminum is dropped into a beaker with 88.2 ml of water. The temperature of the water warms from 35.1 degrees Celsius to 53.7. What quantity of heat energy did the piece of aluminum give off? Specific heat of water = 4.18 J/g*K, the density of water is 1.00 g/mL.

I calculated Delta T (18.6) .... Then added mass of water and mass of aluminum (=132.5g).... and q = 4.18 J/g*C

Then plugged those into:

C = q / (m * delta T) ... C = 0.00167, or ( 1.67 x 10 ^ -3 J)

That was marked HALF Wrong.....^

2) For the reaction, calculate the heat produced by the formation of 15.4 g of KCl.

2KClO3(s) -> 2KCl(s) + 3O2(g) ..... Delta H= -89.4 kJ

I did 15.4 g KCl divided by 74.55 = 0.207 mol

Then set up: -89.4 KJ = 15.4g (0.207)(Delta T) .... Which gave me, -28.

That was marked HALF Wrong.....^

3) If you have a solution which is 2.0 M in A, where A -> B by a first order reaction, and the half-life is 2 hours... What will be [A] in 16 hours?

Completely stumped on this one.

4) Consider this multi-step reaction:

Step 1   Slow   2NO2(g) -> NO3(g) + NO(g)

Step 2   Fast NO3(g) + CO(g) -> CO2(g) + NO2(g)

a. Give the intermediate in this reaction: NO3

b. Give the rate Law for the overall reaction: ?????

c. Give the molecularity of the second step: ?????

Answer 1

1]

Heat lost by Al = Heat gained by water

Q = mCdT

Q = heat

C = heat capacity

dT = Temperatuer change

Heat lost by Al = m*C*dT = 88.2*4.18*18.6 = 6857.3736 J

Heat given off by Al = 6857.3736 J

2]

2KClO3(s) -> 2KCl(s) + 3O2(g)  

From the reaction ,

2 moles of KClO3 produces 2 moles of KCl

MOles of KCl = mass / Molar mass = 15.4 / 74.55 = 0.207

Moles of KClO3 = 0.207

2 moles gives energy of --> -89.4 KJ

0.207 moles gives ---> -89.4*0.207 / 2 = -9.2529 KJ

Answer is -9.2529 KJ

3]

Concetration after n halv lives =[A] / 2^n

here n = 4 (16 years)

[A] after 16 hours = [A] / 2^n = 2 / 2^4 = 0.125 M

[A] = 0.125 M

4]

Net reaction :

NO2 + CO - ----> NO + CO2

slow step is the rate determining step

a]

Intermediate ---> NO3 (which not appears in final step )

b]

Rate = k [NO2]^2

c]

Molecularity = 2 --> bimolecular (since two molecules are involved in the second step )