2 Al + 3 Br2 → 2 AIBr3 Now suppose that 5.6 mol of aluminum reacts with 4.4 mol of bromine. 1. Calculate the mass of aluminum bromide that can be
produced from 5.6 mol of Al. Considering the limiting Bromine.

d. temperate deciduous forest

explanation: no but i do not usually comment

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