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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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3 years ago
Read 2 more answers
FIRST TO ANSWER CORRECTLY GETS BRAINELSIT! pls asap :)
zmey [24]
The protons! Is it the answer
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3 years ago
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If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy __________ L at STP.
Vlad [161]

If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.

<h3>How to calculate volume?</h3>

The volume of a gas at STP can be calculated using the direct proportion method.

According to this question, 50.75 g of a gas occupies 10.0 L at STP, then 129.3g of the same gas will occupy the following:

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Therefore, if 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.

Learn more about volume at: brainly.com/question/12357202

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6 0
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