Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
<em />
To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
Answer: 0.113 moles of NaCl are created as a result of decomposing 12 grams of 
.
Explanation:
To calculate the moles :
The balanced chemical equation for decomposition of is:
According to stoichiometry :
2 moles of give = 2 moles of 
Thus 0.113 moles of give =
of
Thus 0.113 moles of NaCl are created as a result of decomposing 12 grams of .
Its an ore of uraninte i think.
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