What is the correct term for water Atom element product reactant
2 answers:
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1.04 ⨯ 10 M
<em>A</em> = <em>ε</em> by the Beer-Lambert law, where
- <em>A</em> the absorbance,
the path length,
- <em>ε</em> the molar absorptivity of the solute, and
concentration of the solution.
<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, is constant;
is inversely proportional to
. The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.
The 13.0 mL standard solution has a concentration of 5.17 ⨯ M. Diluting it to 45.0 mL results in a concentration of
1.494 M.
is inversely related to
for the two solutions. As a result, c₂ =
3.126 M.
The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of 1.04 ⨯
M.
Hello!
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You'll need to react 7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
</span>
First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):
2Na + H₂SO₄ → Na₂SO₄ + H₂
Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:
Have a nice day!
<span>
You'll need to react 7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
</span>
First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):
2Na + H₂SO₄ → Na₂SO₄ + H₂
Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:
Have a nice day!