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

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

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 the above.

0.50 mol  Na₂CO₃ and 0.50 M Na₂CO₃ are different because:

The numbers 234,000 and 6.0012 can be represented in scientific notation as:

1. $2.34\times {10}^{-9}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>6\times {10}^3$

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

3. $2.34\times {10}^{-9}<mspace\; width="1em"></mspace>\mathrm{and}$ $6.0012\times {10}^{-9}$

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

How many grams of HCl are required to react with 5.0 g of manganese dioxide in the reaction provided?
\(4 \mathrm{HCl}_{(\mathrm{aq})}+\mathrm{MnO}_{2(\mathrm{~s})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}+\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{Cl}_{2(\mathrm{~g})}\)

1. 4.8 g

2. 6.4 g

3. 2.8 g

4. 8.4 g

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

\(CaCO_3 (s)\ + \ HCl (aq)\ \rightarrow CaCl_2(aq)\ + CO_2(g)\ \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ H_2O(l)\)
 

1. 0.36 g

2. 0.09 g

3. 0.96 g

4. 0.66 g

The molar mass of naturally occurring Argon isotopes is: 

The number of significant figures present in the answer of the following calculations [(i), (ii), (iii)]  are respectively -