The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
answer d
Explanation:
methy group is in the second position while the double bond is in the third position
Answer:
3 years
Explanation:
Given data:
Initial amount of sample = 160 Kg
Amount left after 12 years = 10 Kg
Half life = ?
Solution:
at time zero = 160 Kg
1st half life = 160/2 = 80 kg
2nd half life = 80/2 = 40 kg
3rd half life = 40 / 2 = 20 kg
4th half life = 20 / 2 = 10 kg
Half life:
HL = elapsed time / half life
12 years / 4 = 3 years
20 g O2 x 1 mol O2/32 g O = 0.625 mol O2