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lara31 [8.8K]
3 years ago
10

Silver has two naturally occurring isotopes with the following isotopic masses: 47Ag 107 – 106.90509 47Ag 109 – 108.9047 The ave

rage atomic mass of silver is 107.8682 amu. The abundance of the lighter of the two isotopes is __________. 3/1
A) 0.2422B) 0.4816C) 0.5184D) 0.7578E) 0.9047
Chemistry
1 answer:
vovangra [49]3 years ago
3 0

Answer:

c) .51835

Explanation:

Let the relative abundance of the lighter of the two isotopes be X we have

Then the relative abundance of the heavier isotope is then (1-X)

Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.

To calculate the relative atomic mass of silver, we have

(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver

106.90509(X) + 108.9047(1-X)

108.9-108.9(x)+106.9(x) = 107.87

-2x-1.03 = 0.517450902926

Closest answer is c

c) .5184

The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.

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Answer:

4.81 moles

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Pressure at which gauge reads zero = 14.7 psi

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Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

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The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

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Using ideal gas equation as:

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where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

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Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai
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Write the fraction of the mass of kcl produced from 1 g of k2c03​
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Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

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mol of K₂CO₃(MW=138 g/mol) :

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