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Vikki [24]
3 years ago
14

Naturally occurring indium has two isotopes, indium-113(112.9040580 amu) and Indium-115 (114.9038780 amu. The atomic mass of ind

ium is 114.82 amu. What is the percent of indium-115? a)4.32% b)48.4% c)95.7% d)0.973%
Chemistry
1 answer:
Kruka [31]3 years ago
5 0

Answer:

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

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3 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

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