The following data was obtained when dinitrogen and dioxygen react together to form different compounds:

Mass of dinitrogen  Mass of dioxygen
i. 14 g 16 g
ii. 14 g 32 g
iii. 28 g 32 g
iv. 28 g 80 g

The law of chemical combination applicable to the above experimental data is:
1. Law of reciprocal proportions
2. Law of multiple proportions
3. Law of constant composition
4. None of these.

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Round up the following number into three significant figures: 
i. 10.4107  ii. 0.04597 respectively  are 

1. 10.4, 0.0460 2. 10.41, 0.046
3. 10.0, 0.04 4. 10.4, 0.0467
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0.50 mol  Na2CO3 and 0.50 M Na2CO3 are different because:

1. Both have different amounts of  Na2CO3.
2. 0.50 mol is the number of moles and 0.50 M is the molarity.
3. 0.50 mol  Na 2 CO 3  will generate more ions.
4. None of the above.

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The number of significant figures in the numbers 5005, 500.0, and 126,000 are, respectively: 

1. 2, 4, and 3  2. 4, 1, and 3 
3. 4, 4, and 6  4. 4, 4, and 3 
Subtopic:  Introduction |
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The numbers 234,000 and 6.0012 can be represented in scientific notation as -

1. 2.34×10-9 and 6×103

2. 0.234 ×10-6 and 60012×10-9

3. 2.34×10-9 and 6.0012×10-9

4. 2.34×105 and 6.0012×100

Subtopic:  Introduction |
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The correct match is:
List I List II
a. Micro i. 10-15 m
b. Mega ii. 10-6 m
c. Giga iii. 106 m
d. Femto iv. 109 m
 
a b c d
1. i iv iii ii
2. iii iv ii i
3. ii iii iv i
4. i iii iv ii
Subtopic:  Introduction |
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Amount of HCl that would react with 5.0 g of manganese dioxide, as per the
given reaction will be:

4HCl(aq) + MnO2(s) ➡ 2H2O(l) + MnCl2(aq) + Cl2(2)

1. 4.8 g

2. 6.4 g

3. 2.8 g

4. 8.4 g

Subtopic:  Equation Based Problem |
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The mass of CaCO3 required to react completely with 25 mL of 0.75 M HCl according to the given reaction would be:

CaCO3(s) + HCl(aq) ➡ CaCl2(aq) + CO2(g) + H2O(l) 

1. 0.36 g

2. 0.09 g

3. 0.96 g

4. 0.66 g

Subtopic:  Concentration Based Problem | Equation Based Problem |
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The molar mass of naturally occurring Argon isotopes is: 

Isotope Isotopic molar mass Abundance
36-Ar  35.96755 g mol–1 0.337%
38-Ar  37.96272 g mol–1  0.063%
40-Ar  39.9624 g mol–1  99.600%
 
1. 49.99947 g mol-1  2. 39.99947 g mol-1  
3. 35.59947 g mol-1   4. 45.59947 g mol-1  
Subtopic:  Empirical & Molecular Formula |
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The number of significant figures present in the answer of the following calculations [(i), (ii), (iii)]  are respectively -

1. 0 . 02856   ×   298 . 15   × 0 . 112 / 5785
2. 5 × 5.364
3. 0.0125 + 0.7864 + 0.0215
 
1. 4, 4, 3 2. 3, 3, 4
3. 4, 3, 4 4. 3, 4, 4

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