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lara31 [8.8K]
4 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]4 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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​SO2 + 2NaOH → Na2SO3 + H2O
katen-ka-za [31]

Answer:

There will be produced 157.55 grams of Na2SO3. The limiting reactant is NaOH. SO2 is in excess, there will remain 19.9 grams of SO2.

Explanation:

Step 1: Data given

Mass of SO2 = 100 grams

Mass of NaOH = 100 grams

Molar mass of SO2 = 64.07 g/mol

Molar mass of NaOH = 40 g/mol

Molar mass of Na2SO3 = 126.04 g/mol

Step 2: The balanced equation

​SO2 + 2NaOH → Na2SO3 + H2O

Step 3: Calculate moles SO2

Moles SO2 = mass SO2 / molar mass SO2

Moles SO2 = 100.0 grams / 64.07 g/mol

Moles SO2 = 1.561 grams

Step 4: Calculate moles of NaOH

Moles NaOH = 100.0 grams / 40 g/mol

Moles NaOH = 2.5 moles

Step 5: Calculate limiting reactant

For 1 mol of SO2 we need 2 moles of NaOH to produce 1 mol Na2SO3 and 1 mol of H2O

NaOH is the limiting reactant. It will completely be consumed.(2.5 moles).

SO2 is in excess. There will be consumed 2.5 / 2 = 1.25 moles of SO2

There will remain 1.561 - 1.25 = 0.311 moles of SO2. This is 0.311 * 64.07 g/mol = 19.9 grams

Step 6: Calculate moles of Na2SO3

There will be produced 1.25 moles of Na2SO3

Step 7: Calculate mass of Na2SO3

Mass Na2SO3 = 1.25 * 126.04 g/mol

Mass Na2SO3 = 157.55 grams

There will be produced 157.55 grams of Na2SO3. The limiting reactant is NaOH. SO2 is in excess, there will remain 19.9 grams of SO2.

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4 years ago
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Answer:

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