Answer:- 3. 
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is 
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is 
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is . The C to H ratio for second molecule is 1:4, so the empirical formula is 
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also . Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is . As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3. and
Answer:
1 At 0C° KNO3 is least soluble
2 Approximately 65 grams
3 About 30 grams
4 yes it increases at the same rate can be explained by straight line graph
Explanation:
Answer to the Question
C. Barium phosphate
Vanadium
V 1s²2s²2p⁶3s²3p⁶4s²3d³
In one mole of talc, we observe that there are:
3 moles of Mg
4 moles of Si
2 moles of H
12 moles of O
The molar ratio of O to Mg is then:
12 moles of O : 3 moles of Mg = 4 : 1
Therefore, if 6.1 moles of Mg are present, the moles of O are:
4 * 6.1 = 24.4 moles of O
