Chemistry questions and answers

Chemistry Questions and Answers

Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.

426.

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\(_2\)

427.

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.

428.

The products of the thermal decomposition of ammonium trioxonitrate(v) are?

A.

Nitrogen (I) Oxide and Water

B.

Ammonia and Oxide

C.

Nitrogen and Water

D.

Nitrogen (IV) Oxide and Water

Correct answer is A

NH\(_4\)NO\(_3\) Ammonium Trioxonitrate(V) , unlike all other metallic trioxonitrates(v), decomposes on heating to give nitrogen(i)oxide N\(_2\)O and water.

429.

Four elements P, Q, R and S have atomic numbers of 4, 10, 12 and 14 respectively. Which of these elements is noble gas? 

A.

P

B.

Q

C.

R

D.

S

Correct answer is B

They are the most stable due to having the maximum number of valence electrons their outer shell can hold. Therefore, they rarely react with other elements since they are already stable.

430.

If the quantity of oxygen occupying 2.76L container at a pressure of 0.825 atm and 300k is reduced by one-half, what is the pressure exerted by the remaining gas?

A.

1.650atm

B.

0.825atm

C.

0.413atm

D.

0.275atm

Correct answer is A

Using the combined gas law formula,

Given that P\(_1\) = 0.825,   V\(_1\) = 2.76 L, V\(_2\) = 1.38 L, P\(_2\) = ?

And T1 = T2; we would have P\(_2\) = [ P\(_1\) X V\(_1\) ] ÷ V\(_2\)

: [0.825 X 2.76]  ÷ 1.38

= 1.650atm