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jekas [21]
1 year ago
15

when you combust 100.0 g 100.0 g of propane at 500. k 500. k and 1.00 atm 1.00 atm in a closed container, you expected to collec

t 279 l 279 l of carbon dioxide. instead, when you collect the gas, it measures 651 l 651 l in total. which of the following accounts for this difference?
Chemistry
1 answer:
bixtya [17]1 year ago
6 0

for the following accounts Theoretical yield was calculated incorrectly

The random error that results from measuring the masses of the reactants and the product (+ or - 0.001 g?) could cause experimental error in a yield measurement. Or it can be a deliberate mistake brought on by impurities that are still in the final product (and artificially inflating its mass). Because the real yield is frequently lower than the theoretical value, percent yield is typically lower than 100%. This may be due to incomplete or conflicting reactions or sample loss during recovery.

Learn more about Theoretical yield here:

brainly.com/question/22702379

#SPJ4

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4 years ago
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What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor
Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

CH_3COOH(l)\rightarrow CH_3COOH(g)..[1]

Heat of vaporization of acetic acid = H^o_{vap}=24.3 kJ/mol

CH_3COOH(s)\rightarrow CH_3COOH(l)..[2]

Heat of fusion of acetic acid = H^o_{fus}=10.8 kJ/mol

Heat of sublimation of acetic acid = H^o_{sub}=?

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[1] + [2] = [3] (Hess's law)

H^o_{sub}=H^o_{vap}+H^o_{fus}

=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0
3 years ago
What is the molarity of a solution that is made by adding 57.3 g of MgO to 500.0 mL of solution?
kifflom [539]

Answer:

2.843 M

Explanation:

Molarity = moles / volume

<u>Milliliters to liters:</u>

500 mL <u>= .500 L</u>

<u>Grams to moles:</u>

MgO molar mass = 16.00 + 24.31 = 40.31 g/mol

57.3 g x 1 mol / 40.31 g = <u>1.421 mol</u>

<u>Molarity:</u>

1.421 mol / .500 L =  2.843 M

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