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

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

16 g of oxygen has the same number of molecules as in:

(a) 16 g of CO

(b) 28 g of N₂

(c) 14 g of N₂

(d) 1.0 g of H₂

1. (a), (b)

2. (b), (c)

3. (c), (d)

4. (b), (d)

Unitless terms among the following are:

The incorrect statements regarding 1 mole of dioxygen gas at STP from the above four statements are: 

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

1. $2.34\times {10}^{-9} \mathrm{and} 6\times {10}^3$

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

3. $2.34\times {10}^{-9} \mathrm{and}$ $6.0012\times {10}^{-9}$

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

0.50 mol  Na₂CO₃ and 0.50 M Na₂CO₃ are different because:

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