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Sergio [31]
3 years ago
15

What mass of sodium oxalate na2c2o4 is needed to prepare 0.250 l of a 0.100 m solution?

Chemistry
2 answers:
Lena [83]3 years ago
7 0
Molarity = number of moles of solute / liters of solution
number of moles = molarity * liters of solution
number of moles of Na2C2O4 = 0.1 * 0.25 = 0.025 moles

Now, from the periodic table:
mass of Na = 22.9 grams
mass of C = 12 grams
mass of O = 16 grams
molar mass of Na2C2O4 = 2(22.9) + 2(12) + 4(16) = 133.8 grams
Therefore, one mole is equal to 133.8 grams. To know the mass of 0.025 moles, all you have to do is cross multiplication as follows:
mass = (0.025*133.8) / 1 = 3.345 grams
Nataly_w [17]3 years ago
6 0

Answer:

3.35 g Na2C2O4

Explanation:

They gave us the molarity and the liters of solution so using the relationship (M=mol/V) we can get the moles of solution.

0.250 L (0.100 mol/L) = 0.0250 mol Na2C2O4

Now that we have the moles of sodium oxalate, we can use its molar mass (134 g/mol) to convert to grams.

0.0250 mol (134.0 g/mol) -=3.35 grams (rounded to three sig figs

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Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

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pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

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-0.644 = Log [A⁻]/[HA]

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4.87 = 4.874 + Log [A⁻]/[HA]

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  • (c)

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0.426 = Log [A⁻]/[HA]

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2.667 = [A⁻]/[HA]

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4 years ago
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3 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
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Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

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The balanced equation for the reaction is,
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Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
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