Answer:
2
Explanation:
The sample has halved twice.
1/2 of 1/2 is 1/4, and 1/2 of 1 is 1/2.
The percent yield for this reaction is 91%.
<u>Explanation</u>:
Ba(NO3)2 + Na2SO4 — > BaSO4 + 2NaNO3
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Assume that 0.45 mol of Ba(NO3)2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4.
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Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield.
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Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)
= 96 g BaSO4
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Beginning with 0.45 mol of Ba(NO3)2, compute the theoretical yield of BaSO4. Mole proportion of Ba(NO3)2 to BaSO4 is 1:1.
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0.45 mol Ba(NO3)2 x (1 mol BaSO4/1 mol Ba(NO3)2) x (233.38 g BaSO4/1 mol BaSO4) = 105 g BaSO4
% Yield = (Actual Yield/Theoretical Yield) x 100%
= (96/105) x 100%
= 91%.
Answer:
77.5°C
Explanation:
Data obtained from the question include the following:
Mass (M) = 100g
Heat (Q) = 1x10^3 J
Change of temperature (ΔT) =..?
Note: The specific heat capacity (C) of gold = 0.129 J/g°C
With the above information obtained from the question, we can obtain the change in temperature as follow:
Q = MCΔT
Divide both side by MC
ΔT = Q/MC
ΔT = 1x10^3 / (100 x 0.129)
ΔT = 77.5°C
Therefore, the change in temperature is 77.5°C
Answer:
C Is the correct answer!
Took the Cumulative Exam Review :/
