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lina2011 [118]
3 years ago
8

Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass

of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.
Chemistry
1 answer:
Yanka [14]3 years ago
7 0
You will have to do some math. 

<span>80.2 × 11.01 + 19.8 × x = 100 * 10.81. </span>

<span>x is the mass in amu of the second isotope. </span>

<span>Solve for x. </span>

<span>19.8 × x = 100 * 10.81 - 80.2 × 11.01 </span>
<span>19.8 × x = 1081 - 883.00 = 198.00 </span>
<span>x = 1964.00 / 19.8 = 10.00 </span>

<span>The mass of the other isotope is 10.00 amu.</span>
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Answer:

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Explanation:

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2 years ago
Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5
siniylev [52]

Answer:

10^{-3.4

Explanation:

Let us first take a look at the image below;

In the acid - base  reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

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∴ The equilibrium constant for the acid-base reaction is expressed as:

K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}

      = \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}

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From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}

K_{eq}=\frac{10^{-5.8} of hydrobromic acid}{10^{-2.4} of Ethyloxonium acid}

       = 10^{-3.4}

6 0
3 years ago
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3 years ago
8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4
Klio2033 [76]

<em>ANSWER</em>

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STEP-BY-STEP EXPLANATION:

Given information

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To calculate the number of moles, we will be using the below formula

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Recall that, the Avogadro's constant is given as

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Therefore, the number of moles of methane is 905.32 moles

6 0
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