Answer: <span>option C. Measure the four products separately, but using the
same scale each time.
</span>
Justification:
Review the choices given one by one.
<span> A. Measure the product separately
on four different scales.
</span><span>
</span><span>
</span><span>He should not include other variables in the experiment. Using four different scales migh give different results due to differences on the scales.
</span><span>
</span><span>B. Measure all of the product together and
divide by four.
</span><span>
</span><span>
</span><span>By doing this he will just obtain an average of the amount produced, but in this way he cannot determine the production individually (with each catalyst).
</span><span>
</span><span>
</span><span>C. Measure the four products separately, but using the
same scale each time.
</span><span>
</span><span>
</span><span>Indeed doing this he will be able to compare and rank the efficiency of each catalyst.
</span><span>
</span><span>
</span><span>D. Measure the amount of sodium chloride in one
sample every 10 seconds.
</span>
He cannot do this without affecting at the same time the evolution of the reaction, and of course by doing it on one sample only he will not be able to compare.
Answer:
Explanation:
The most easily reduced groups in a protein are disulfide bonds, RS-SH.
Answer:
96.0 grams of oxygen gas (O2) will be produced.
Option B is correct.
Explanation:
Step 1: Data given
Number of moles SO2 = 2.0 moles
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
Step 3: Calculate moles oxygen gas (O2)
For 2 moles H2S we need 3 moles O2 to produce 2 moles SO2 and 2 moles H2O
For 2.0 moles SO2 produced we need 3/2 * 2.0 = 3.0 moles O2
Step 4: Calculate mass O2
Mass O2 = moles O2 * molar mass O2
Mass O2 = 3.0 moles * 32.0 g/mol
Mass O2 = 96.0 grams
96.0 grams of oxygen gas (O2) will be produced.
Option B is correct.
