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3 years ago

What is the sum of the numbers of neutrons and electrons in the ion 208 Pb2+?

Chemistry

1 answer:

AysviL [449]3 years ago

3 0

Answer:

For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206

Explanation:

Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.

The isotope of lead with the mass number 208, has 126 neutrons.

Since, atomic number = number of protons = number of electrons for neutral atom

Therefore, for ²⁰⁸Pb: number of electrons= 82

So, <u>for ²⁰⁸Pb²⁺ cation: number of electrons</u>= 82 - 2 = 80

<u>Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.</u>

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3 years ago

An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t

Elenna [48]

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

$C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+$

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

$n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol$

Thus, the concentration of ethylamine in solution is:

$[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M$

Now, we can also infer that some salt is formed, and has the following concentration:

$[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M$

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

$pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0$

Finally, the pH turns out to be:

$pH=14-pOH=14-3\\\\pH=11$

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

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3 years ago

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Examples are Iron exsits as Fe+2 and Fe+3

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3 years ago