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Sedbober [7]
3 years ago
13

A new compound was recently discovered and found to have an atomic weight of 342.38 amu. This element has two isotopes, the ligh

ter of which has a mass of 340.91 amu and an abundance of 68.322%. What is the mass of the heavier isotope?
Chemistry
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer:

The mass of heavier isotope is 345.6 amu.

Explanation:

Given data:

atomic wight of compound = 342.38 amu

lighter isotope mass = 340.91 amu

abundance of lighter isotope = 68.322%

mass of heavier isotope = ?

Solution:

average atomic mass = ( % age abundance of lighter isotope × its atomic mass) + (% age abundance of heavier isotope × its atomic mass)  / 100

percentage of heavier isotope = 100- 68.322 = 31.678

Now we will put the values in formula.

342.38  = (68.322× 340.91) + (31.678 × X) / 100

342.38 = 23291.65302 +  (31.678 × X) / 100

342.38 × 100 = 23291.65302 +  (31.678 × X)

34238 -23291.65302 =  (31.678 × X)

10946.35 / 31.678 = X

345.6 = X

The mass of heavier isotope is 345.6 amu.

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8 0
3 years ago
In the citric acid cycle, malate is dehydrogenated to oxaloacetate in a highly endergonic reaction with a ΔG’o of +30 kJ mol-1:
Doss [256]

Answer :  The value of K_{eq} of this reaction is, 5.51\times 10^{-6}

At equilibrium, [L-malate] > [oxaloacetate]

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = +30 kJ/mol = +30000 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

The given reaction is:

\text{L-malate}+NAD^+\rightleftharpoons \text{oxaloacetate}+NADH+H^+

\Delta G^o=-RT\times \ln K_{eq}

+30000J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=5.51\times 10^{-6}

Therefore, the value of K_{eq} of this reaction is, 5.51\times 10^{-6}

As, the value of K_{eq} < 1 that means the reaction mixture contains reactants.

At equilibrium, [L-malate] > [oxaloacetate]

7 0
4 years ago
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