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.

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

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 these.

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

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$\times {10}^0$

Amount of HCl that would react with 5.0 g of manganese dioxide, as per the
given reaction will be:
4HCl_(aq) + MnO_2(s) ➡ 2H₂O_(l) + MnCl_2(aq) + Cl₂₍₂₎

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) ➡ CaCl_2(aq) + CO_2(g) + H₂O_(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 -