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oksano4ka [1.4K]
4 years ago
9

What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C

2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)
Chemistry
1 answer:
Greeley [361]4 years ago
5 0

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

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Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?
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An alkali metal present in period 2 have larger first ionization energy.

Explanation:

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The amount of energy required to remove the electron from the atom is called ionization energy.

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\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

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