WAEC Past Questions and Answers - Page 1404

7,016.

Study the table below and use it to answer the question

 

 Cost Price (N)  Useful Life (Life)  Salvage value (N)  Annual depreciation (N)  Total depreciation
 20000  Y  5000  1000 annum  15000
 60000  10  18000  4200 annum  Z

The useful life represented by Y in the illustrated table is

A.

3 years

B.

4 years

C.

15 years

D.

20 years

Correct answer is C

Cost Price (N) = 20000
Salvage value (N) = 5000
Annual depreciation (N) = 1000
Total depreciation = 15000
The formula to calculate the useful life is:
Useful life = (Cost Price - Salvage value) / Annual depreciation
Now, let's plug in the values:
Useful life = (20000 - 5000) / 1000
Useful life = 15000 / 1000
Useful life = 15 years
Therefore, the useful life represented by Y in the table is 15 years.

7,017.

What is the expression for the equilibrium constant (Kc) for the following reaction? N2(g) + O2(g) ⇌ 2NO(g)

A.

\(\frac{[NO^2]}{[N^2] + [O^2]}\)

B.

\(\frac{[2NO]^2}{[N_2][O_2]}\)

C.

\(\frac{[N_2] + [O_2]}{2[NO]^2}\)

D.

\(\frac{[NO]^2}{[N_2][O^2]}\)

Correct answer is D

No explanation has been provided for this answer.

7,018.

What will happen if more heat is applied to the following system at equilibrium?

X\(_2\)(g) + 3Y\(_2\)(g) ⇌ 2XY\(_3\)(g); ∆H = xKJmol\(^{-1}\)

A.

the yield of XY 3 will decompose

B.

more of XY3 will decompose

C.

More of X2 will react

D.

the forward reaction will remain static

Correct answer is C

Increase in temperature will cause the forward reaction to occur, increasing the amounts of the products and decreasing the amounts of reactants.

7,019.

The bacterium which is responsible for conversion of ammonium compounds to nitrites in the nitrogen cycle is

A.

Nitrobacter

B.

Rhizobium

C.

Nitrosomonas

D.

Azotobacter

Correct answer is B

Nitrosomonas is the bacterium responsible for the first step in this conversion process, where it oxidizes ammonium compounds to nitrites. This conversion is essential as it prepares the nitrogen compounds for the subsequent steps in the nitrogen cycle.

7,020.

Calculate the mass of sodium hydroxide in 5.00 dm3 of a 0.125 mol dm-3 solution. [NaOH = 40 g mol-1]

A.

0.0156 g

B.

0.625 g

C.

1.00 g

D.

25.0 g

Correct answer is D

1 mole of NaOH per dm3 has a mass of 40g.
mass of NaOH in 0.125mole
= 0.125 x 40 = 5g.
mass in 5.00 dm3 of the solution.
= 5 x 5 = 25.0g.