The concentration of the sodium chloride would be 0.082 M
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.
Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles
Equivalent mole of NaCl = 0.325 moles.
Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M
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Answer:
To calculate an electron configuration, divide the periodic table into sections to represent the atomic orbitals, the regions where electrons are contained. Groups one and two are the s-block, three through 12 represent the d-block, 13 to 18 are the p-block and the two rows at the bottom are the f-block.Explanation:
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
True because they are organically made without any things that include chemicals
Answer:
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