If you want to make 300.0 grams of AgBr, how many moles of HBr will you need? Use the balanced chemical equation. Ag + 2 HBr → 2 AgBr + H₂

b. is the answer hoped i        

explanation:

0.07 g/cm³

1.37 g/cm³

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