You place 1.75 mol of pure in an 8.0-L flask at 1000 K. At equilibrium, 0.23 mol of has been formed. Calculate for the reaction at 1000 K.

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

answer carbon and three oxygen atoms;

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally