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
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
Answer:
72.53% is the yield of CrCl3
Explanation:
Given
Reaction:
Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)
CCl4 is in excess and 17.6g Cr2O3 present
The reaction yields 26.6g of CrCl3
To Find:
% yields of the reaction
Also given
Molar mass of CrCl3 = 158.35g/mol
Molar mass of Cr2O3 = 152.00 g/mol
By the stoichiometry of the reaction
1 mole of Cr2O3 gives 2 moles of CrCl3
0r
1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3
= 1 52 g of Cr2O3 gives 316.70 g of CrCl3
17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3
= 36.67 g CrCl3
but actual yield is only 26.6g
so % yield is (26.6 ÷÷ 36.67) × 100
= 72.53% is the yield of CrCl3
It is 0.5474 but you can put 0.5 hope this helps
