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Hunter-Best [27]
3 years ago
11

There are two isotopes of boron: boron-10 and boron-11. Explain, in terms of weighted average atomic mass, the difference in per

cent abundance of each of the isotopes of boron.
Chemistry
1 answer:
Zolol [24]3 years ago
8 0

Answer:

Percent abundance of  B¹⁰ is 19.9% and  B¹¹ is 80.1%.

The weighted average atomic mass of boron is 10.811 amu.

Explanation:

We know that average atomic mass of Boron is 10.811 amu and there are two isotopes of boron B¹⁰ (10.012938) and B¹¹ (11.009305)

We will determine the percent abundance of each isotopes.

First of all we will set the fraction for both isotopes B¹⁰ and  B¹¹.

X for the isotopes having mass 11.009305 amu.

1-x for isotopes having mass 10.012938 amu.

The average atomic mass of Boron is 10.811 amu

we will use the following equation,

11.009305x +  10.012938  ( 10.012938 -x) = 10.811

11.009305x +  10.012938 -  10.012938 x = 10.811

11.009305x   -  10.012938 x = 10.811 -  10.012938

0.996367x = 0.798062

x= 0.798062 / 0.996367

x= 0.800972

Percent abundance of B¹¹.

0.800972 × 100 = 80.1 %

80.1 % is abundance of B¹¹ because we solve the fraction x.

now we will calculate the abundance of B¹⁰.

(1-x)

1-0.800972 =0.199

Percent abundance of  B¹⁰.

0.199 × 100=  19.9%

19.9% for B¹⁰.

Now we can calculate the average atomic mass of boron.

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [19.9 × 10.012938] + [80.1 × 11.009305] / 100

Average atomic mass = 199.2575 + 881.8453 / 100

Average atomic mass = 10.811 amu

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Explanation:

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Now, to solve this problem, we use the combining powers which corresponds to the number of electrons usually lost or gained or shared by atoms during the course of a chemical combination.

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Now the molecular formula is  Pb₂O₄

                               

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What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

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For some reversible reactions that involve gases:

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Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

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The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

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For reactions that involve gases,

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However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

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. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub
elena-s [515]

The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

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  • d = m/v

Where:

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  • v(i) = initial volume

Information about the problem:

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Applying the volume formula we get:

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v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

Learn more about density in: brainly.com/question/1354972

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