# If the speed of light is 3.0 × 108 m s–1, then the distance covered by light in 2.00 nanoseconds will be - 1. 0.500 m 2. 0.600 m 3. 0.700 m 4. 0.800 m

Subtopic:  Introduction |
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In a reaction A + B2 → AB2,
A will act as a limiting reagent if :

 i. 300 atoms of A reacts with 200 molecules of B ii. 2 moles A reacts with 3 moles B2 iii. 100 atoms of A reacts with 100 molecules of B iv. 5 moles A reacts with 2.5 moles B v. 2.5 moles A reacts with 5 moles B2

Choose the correct option:

1. (i), (ii)
2. (i), (ii), (v)
3. (ii), (v)
4. All

Subtopic:  Limiting Reagent |
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If 10 volumes of H2 gas react with 5 volumes of O2 gas, the volumes of water vapor produced would be:

1. 9

2. 8

3. 10

4. 11

Subtopic:  Equation Based Problem |
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15.15 pm in the basic unit will be

1. 1.515 × 1012 m

2. 2.57 × 1011 m

3. 2.87 × 1011 m

4. 1.515 × 10–11 m

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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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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 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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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 correct match is:
 List I List II a. Micro i. ${10}^{-15}$ m b. Mega ii. ${10}^{-6}$ m c. Giga iii. ${10}^{6}$ m d. Femto iv. ${10}^{9}$ m
Codes:
 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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The numbers 234,000 and 6.0012 can be represented in scientific notation as:

1.

2. 0.234 $×{10}^{-6}$ and $60012×{10}^{-9}$

3.  $6.0012×{10}^{-9}$

4. 2.34$×{10}^{5}$ and 6.0012$×{10}^{0}$

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