Researchers conducted experiments with trees. Listed below are weights (kg) of t

Question

Researchers conducted experiments with trees. Listed below are weights (kg) of trees given no fertilizer and trees treated with fertilizer and irrigation. Find the range and standard deviation for each of the two samples, then compare the two sets of results. Does there appear to be a difference between the two standard deviations? Find the range for the trees that were given no treatment. Range = kg Find the range for the trees treated with fertilizer and irrigation. Range = kg Find the sample standard deviation for the trees that were given no treatment Sample standard deviation = kg Now find the sample standard deviation for the trees treated with fertilizer and irrigation. Sample standard deviation = kg is there a difference between the two standard deviations? A. Yes, the treated trees appear to have weights that vary much less than the weights of the trees with no treatment. B. Yes, the trees with no treatment appear to have weights that vary much less than the weights of the treated trees. C. No, there is not a difference the two standard deviations as both values are very small.

Explanation / Answer

A)

Cu (s) + 4 HNO3 (aq) ---------------------------> Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)

B)

moles of copper = 0.1059 / 63.546

                          = 0.00167

Cu (s) + 4 HNO3 (aq) ---------------------------> Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)

from balanced equation

1 mole of Cu --------------------- 4 moles HNO3

0.00167 mole Cu -------------- 0.00167 x 4 moles HNO3

= 0.00667 mole HNO3

moles of HNO3 = 0.00667

molarity = 9.0 M

molarity = moles / volume

9.0 = 0.00667 / volume

volume = 0.000741 L

volume = 0.741 mL

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