Naturally occurring indium has two isotopes, indium-113(112.9040580 amu) and indium-115 (114.9038780 amu. the atomic mass of indium is 114.82 amu. what is the percent of indium-115? a)4.32% b)48.4% c)95.7% d)0.973%

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu)  ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂)  / 100

%₁ = x I established this is an equation

%₂ = 100 - x

Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100    and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x  /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

it would be a gas formation

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2