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

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$\times {10}^0$

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

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 

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.

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 -