How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?
2 answers:
Answer : The volume of needed are, 0.013 L
Explanation :
According to the neutralization law,
where,
= molarity of
= 6.0 M = 6.0 mole/L
= molarity of
= 2.0 M = 2.0 mole/L
= volume of
= ?
= volume of
= 39 ml = 0.039 L
conversion : 1 L = 1000 ml
Now put all the given values in the above formula, we get the volume of
Therefore, the volume of needed are, 0.013 L
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Explanation:
For the first part,
Reaction equation:
N₂ + 3H₂ → 2NH₃
Given:
Number of moles of NH₃ = 6 moles
Unknown:
Number of moles of N₂ = ?
Solution:
N₂ + 3H₂ → 2NH₃;
From the reaction above, we solve from the known specie to the unknown. Ensure that the equation is balanced;
2 moles of NH₃ is produced from 1 mole of N₂
6 moles of NH₃ will be produced from mole of N₂
= 3moles of N₂
The number of moles of N₂ is 3 moles
ii.
Given parameters:
Number of moles of sulfur = 2.4moles
Molar mass of sulfur = 32.07g/mol
Unknown:
Mass of sulfur = ?
Solution:
The number of moles of any substance can be found using the expression below;
Number of moles =
Mass of sulfur = number of moles of sulfur x molar mass
Insert the parameters and solve;
Mass of sulfur = 2.4 x 32.07 = 76.97g
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.
