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ch4aika [34]
3 years ago
5

What is the answer to this ?

Chemistry
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

12.78 or 12.80 mostly like 12.78

Explanation:

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In a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes. Determine the average relative atomic
Burka [1]

The average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

<h3>What is Atomic mass? </h3>

Atomic mass is defined as the whole mass of an atom.

It is also defined as the sum of atomic number and number of neutrons.

Atomic mass = Atomic number + neutrons

<h3>What is Isotopes?</h3>

Isotopes are the those element which have same atomic number but have different mass number and number of neutrons.

The average relative atomic mass can be calculated as

mass of isotopes/ mass of sample

mass of all isotopes = (70 × 9X) + (30 × 11X)

=(630 + 330) X

= 960X

Average relative atomic mass = 960X/ 100 X

= 9.6 g

Thus, we concluded that the average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

learn more about atomic mass:

brainly.com/question/14250653

#SPJ9

5 0
1 year ago
What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

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