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LiRa [457]
3 years ago
6

A particular element exists in two stable isotopic forms. One isotope has a mass of 78.9183 amu (50.69% abundance). The other is

otope has a mass of 80.9163 amu (49.31% abundance). Calculate the average mass of the element and determine its identity.
Chemistry
1 answer:
marusya05 [52]3 years ago
5 0

Answer:

Average atomic mass = 79.9035 amu.

The given element is bromine.

Explanation:

Given data:

Mass of 1st isotope = 78.9183 amu

Percent abundance of 1st isotope = 50.69%

Mass of 2nd isotope = 80.9163 amu

Percent abundance of 1st isotope = 49.31%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50.69×78.9183 )+(49.31× 80.9163) /100

Average atomic mass =  4000.3686+ 3989.9828 / 100

Average atomic mass  = 7990.3514 / 100

Average atomic mass = 79.9035 amu.

The given element is bromine.

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Explanation:

D(aq) + E(aq) <=> F(aq)

This question is based on Le Chatelier's principle.

Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Increase D

D is a reactant. if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

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E is a reactant. if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Increase F

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no shift in the direction of the net reaction, Both changes cancels each other.

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Sveta_85 [38]
IAL is the suffix pertaining to conditions percent are birth.
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Answer:

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Maksim231197 [3]

The theoretical yield of SnS_2 will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of SnBr_4 to Na_2S is 1:2.

Mole of 48.1 mL, 0.478 M  SnBr_4 = 0.478 x 48.1/100 = 0.023 mols

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SnBr_4Na_2S is the limiting reactant.

Mole ratio of  SnBr_4  and SnS_2 = 1:1

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More on yields of reactions can be found here: brainly.com/question/17042787

#SPJ1

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

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