Answer:
The Net reaction is
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Explanation:
From the Question we are told that the buffers are
and 
When NaOH is added the Net ionic reaction would be
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Answer:
There are 17.64% students received B+ grades.
Explanation:
It is given that,
Total number of students in chemistry class is 17
We need to find the percentage received by B+.
Number of students having B+ grades are 3 (from graph)
Required percentage = 
So, there are 17.64% students received B+ grades.
The semi will have the hardest time changing direction because of its mass. The more mass there is, the more effort it takes to accelerate and decelerate as well as change direction.
<span>According to the question-
1 mol C3H8O = 60.096 g C3H8O
2 mol C3H8O = 9 mol O2
1 mol O2 = 31.998 g O2
[(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2
Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess.
Next, we need to convert 0.199565 g of O2 into moles of O2:
[(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>