All you have to do is a simple division "parts over the whole"
12.3 grams H/ 110 grams compound x 100= 11.2%
Answer:
c.boron-11
Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.81u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u=ux
x=0.19
1−x=0.81
And thus the abundance of boron-11 is roughly 81%.
Answer:
Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius. ... This results in a larger atomic radius.
Explanation:
Now lets d8
Answer:
Substances 1 and 2
Explanation:
an element only has 1 kind of atoms :3
Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>
In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.
See figure 1
I hope it helps!