D) Radioactive contamination
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
<span>Of the answers listed option B looks like the most complete. Ie "Check for the presence of alpha, beta, and gamma particles." the significant presence of these particles is a specific indicator of radioactive decay, i.e: unstable atoms spontaneously undergoing a nuclear reaction.</span>
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!