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geniusboy [140]
3 years ago
13

How many neutrons do (k) potassium-39 and potassium-40 have, respectively?

Chemistry
1 answer:
Maurinko [17]3 years ago
7 0
Potassium’s atomic number in periodic table is 19, Potassium can only have a mass numbers of 41, 40 and 39. Potassium 39, 40, and 41 are all different isotopes of the element potassium. Each of the isotopes has 19 electrons and 19 protons but the number of neutrons vary. So, in your question Potassium-39 has twenty (20) neutrons, and potassium-40 has 21 neutrons.
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According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

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Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

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