JAMB Past Questions and Answers - Page 1245

6,221.

Which of the following gases will rekindle a brightly splint?

A.

NO\(_2\)

B.

NO

C.

N\(_2\)O

D.

CL\(_2\)

Correct answer is C

Nitrogen(I)Oxide rekindles a brightly glowing splinter but extinguishes a feebly glowing one which is not hot enough to decompose the gas to liberate oxygen

6,222.

The volume occupied by 1.58g of a gas at S.T.P is 500cm\(^3\). What is the relative molecular mass of the gas?  [ G.M.V at S.T.P = 22.4dm\(^{3}\) ]

A.

28

B.

32

C.

44

D.

71

Correct answer is D

500cm\(^3\) of the gas weighed 1.58

: 22400cm\(^3\) of the gas weight

 

[22400 X 1.58g] ÷ [ 500 ]

= 71

6,223.

Work-in-progress 1/1........................N1,000
Work-in-progress 31/12......................N2,000
Production cost of goods manufactured.......N20,000
Sales.......................................N50,000
Stock of finished goods 1/1.................N4,000
Stock of finished goods 31/12...............N5,000
Selling and distribution expenses...........N2,000
Administrative expenses.....................N1,000

Determine the gross profit?

A.

49,000

B.

48,000

C.

31,000

D.

30,000

Correct answer is C

    Trading profit and Loss Account 

opening stock                   4,000
Add: cost of products       20,000      24,000
Less: closing stock                            5000
                                                         19,000

Gross profit                                      31,000
                                                         50,000
selling & dis. Exp               2000
Admin. Exp                       1,000         3000
Net profit                                           28,000

                                                         31,000             

Sales             50,000







G.profit b/d    31000


 

 

 

6,224.

The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO\(_3\) is treated with 0.2dm\(^3\) of 1 Mole of HCL in the equation

CaCO\(_3\) + 2HCL ⇒  CaCl\(_2\)  + H\(_2\) O + CO\(_2\)  , is?

A.

1.00 X 10\(^{23}\)

B.

6.02 X 10\(^{23}\)

C.

6.02 X 10\(^{22}\)

D.

6.02 X 10\(^{24}\)

Correct answer is C

1 mole of CaCO\(_3\) = 100g or  6.02 X 10\(^{23}\)

liberated 1 mole of CO\(_2\) or 44g or 6.02 X 10\(^{23}\)

 

:100g =  6.02 X 10\(^{23}\)

10g of CaCO\(_3\) = x

cross multiply

[10g X 6.02 X 10\(^{23}\)] / 100g = x

⇒ x = 6.02 X 10\(^{22}\)

 

Since 

 6.02 X 10\(^{23}\) of CaCO\(_3\) liberated 6.02 X 10\(^{23}\) of CO\(_2\)

 

⇒ 6.02 X 10\(^{22}\) of CaCO\(_3\) will liberate 6.02 X 10\(^{22}\) of CO\(_2\)

6,225.

How many valence electrons are contained in the element \(^{31}_{15}P\)?

A.

3

B.

4

C.

15

D.

31

Correct answer is A

The element is Phosphorus in group 5 with valencies of 3 and 5.