# 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

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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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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
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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  Na2CO3  will generate more ions. 4 None of the above.

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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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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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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

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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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