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AlexFokin [52]
2 years ago
9

Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith

ium found on the periodic table to determine which isotope is more abundance
Chemistry
1 answer:
alexandr402 [8]2 years ago
4 0

The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

#SPJ1

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<u>Answer:</u>

<u>For a:</u> The balanced equation is 2S(s)+3O_2(g)\rightarrow 2SO_3(g)

<u>For c:</u> The balanced equation is 2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

  • <u>For (a):</u>

The given unbalanced equation follows:

S(s)+O_2(g)\rightarrow SO_3(g)

To balance the equation, we must balance the atoms by adding 2 infront of both S(s) and SO_3 and 3 in front of O_2

For the balanced chemical equation:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

  • <u>For (b):</u>

The given balanced equation follows:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

The given equation is already balanced.

  • <u>For (c):</u>

The given unbalanced equation follows:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)

To balance the equation, we must balance the atoms by adding 2 infront of H_2O(l)

For the balanced chemical equation:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

  • <u>For (d):</u>

The given balanced equation follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

The given equation is already balanced.

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The element in Group 14 Period 4 has how many neutrons and is it a metal, nonmetal, metalloid or noble gas.
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Number of neutrons=41, and it is metalloid

<h3>Further explanation</h3>

Group 14⇒valence electron = 4(ns²np²)

Period 4⇒n=4

So electron configuration of the element :

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While the mass number: 73

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