Absolute zero that means cooling - 0 degree
C6H6(l) + 6 Cl2(g) = C6Cl6(s) + 6 HCl(g)
Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Technically speaking, yes you can. Using a microscope though.
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s