Upon reaction of 1.27 g of copper sulfate with excess zinc metal, 0.416 g copper metal was obtained as an experimental result according to the equation: CusOlaq) + Zn(s) Cu(s) + ZNSO(aq) Before the experiment was carried out the theoretical yield calculation was completed to predict the maximum amount of copper that could be isolated. The answer was 0.506 g of Cu. What is the percent yield in standard notation? [report your answer to the tenths place, i.e. one decimal place.] Type your answer

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Upon reaction of 1.27 g of copper sulfate with excess zinc metal, 0.416 g copper metal was obtained as an experimental result according to the equation:
CuSO (aq) + Zn(s)
→ Cu(s) + ZNSO4(aq)
Before the experiment was carried out the theoretical yield calculation was completed to predict the maximum amount of copper that could be isolated. The answer was 0.506 g
of Cu.
What is the percent yield in standard notation?
[report your answer to the tenths place, i.e. one decimal place.]
Type your answer
Transcribed Image Text:Upon reaction of 1.27 g of copper sulfate with excess zinc metal, 0.416 g copper metal was obtained as an experimental result according to the equation: CuSO (aq) + Zn(s) → Cu(s) + ZNSO4(aq) Before the experiment was carried out the theoretical yield calculation was completed to predict the maximum amount of copper that could be isolated. The answer was 0.506 g of Cu. What is the percent yield in standard notation? [report your answer to the tenths place, i.e. one decimal place.] Type your answer
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