How many moles of calcium nitrate would react with 4.55 moles of chromium (III) sulfate to produce calcium sulfate and chromium(III) nitrate?

13.65 moles

Explanation:

The equation of the reaction is;

3Ca(NO3)2 + Cr2(SO4)3 = 3Ca(SO4) + 2 Cr(NO3)3

4.55 moles of chromium (iii) sulfate reacts with calcium nitrate

The mole ratio of Chromium (iii) sulfate to calcium nitrate is 1: 3

Therefore; the moles of calcium nitrate will be given by

 = (4.55/ 1)×3

= 13.65 moles

Hence; 13.65 moles of calcium nitrate would react with 4.55 moles of Chromium (iii) sulfate.

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