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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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Answer:

\boxed {\boxed {\sf 2.8 \ m }}

Explanation:

The formula for molality is:

m=\frac{moles \ of \ solute}{kg \ of \ solvent}

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

moles= 0.210 \ mol \\kilograms = 0.075 \ kg

Substitute the values into the formula.

m= \frac{ 0.210 \ mol }{0.075 \ kg}

Divide.

m= 2.8 \ mol/kg= 2.8 \ m

The molality is <u>2.8 moles per kilogram</u>

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∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

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⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

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b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

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