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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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Evgen [1.6K]
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6 0
3 years ago
For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

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