Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.
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.
Answer:
The diameter of the hydrogen 
Explanation:
From the given information:
Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

Where the generic expression for angular momentum is:
L = mvr.
replacing the value of L into the previous equation, we have:

----- (1)
The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.
----- (2)
replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:



For ground-state n = 1






Therefore, the diameter of hydrogen d = 2r

