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zzz [600]
3 years ago
10

Indium has only two naturally occurring isotopes. the mass of indium-113 is 112.9041 amu and the mass of indium-115 is 114.9039

amu . part a use the atomic mass of indium to calculate the relative abundance of indium-113.
Chemistry
1 answer:
brilliants [131]3 years ago
8 0
The average atomic mass written in the periodic table is determined from the relative abundances of the element's isotopes. The equation would be:

Average Atomic Mass = ∑(Relative Abundance×Mass)

Let be the relative abundance of Indium-113. Because there are only 2 isotopes, their relative abundances should equal to 1, such that the relative abundance for Indium-15 is (1-x). The atomic mass of indium is 114.818 amu.

x(112.9041) + (1 - x)(<span>114.9039) = 114.818
Solving for x,
x = 0.043 or 4.3%</span>
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B) 12.9 grams.

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(The volume of that one mole of gas at STP will be 22.7 liters if STP is defined as 0 °C and 10⁵ Pa).

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The coefficient in front of Na in the equation is twice the coefficient in front of H₂. It takes two moles of Na to produce one mole of H₂.

n(\text{Na}) = 2\;n(\text{H}_2) = 2\times 0.28125= 0.5625\;\text{mol}.

What's the mass of that many Na atoms?

Refer to a modern periodic table. The molar mass of ₁₁Na is 22.990. The mass of one mole of Na atoms is 22.990 gram. The mass of 0.5625 moles of Na atoms will be

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