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jok3333 [9.3K]
3 years ago
5

After undergoing alpha decay, an atom of radium-226 becomes

Chemistry
1 answer:
Zanzabum3 years ago
8 0
The answer is b. radon-222. The alpha decay means that it will emit an alpha particle when decays. The alpha particle has two protons and two neutrons. So Radium(88) minus two protons will become Radon(86). And the atomic mass will become 226-4=222.
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Complete this equation for the dissociation of na2co3(aq). omit water from the equation because it is understood to be present.
Elanso [62]
Note that we are omitting the water.

So, sodium carbonate will basically dissociate into positive sodium ions and negative carbonate ions based on the following equation:
 <span>Na2CO3 → 2 Na(+) + CO3(2-)
</span>
If we took water into consideration:
Sodium carbonate will dissociate in water forming carbonic acid and sodium hydroxide. Since sodium hydroxide is a strong base, therefore, it will then neutralize the gastric acid, thus, acting as an antacid.
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How to tell which acid whill have the hightest perceent ionization
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We can tell which acid which have the highest percent ionization on the basis of pH value.

<h3>What is percent ionization?</h3>

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2 years ago
What is the ph of a (6.00x10^-3) m hi solution?
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3 years ago
Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O
Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0
3 years ago
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