Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
8.........................................................................................
Answer:
There are 2.258 x 10^24 molecules of CO2 in 165 grams of CO2.
Explanation:
First you have to calculate the molar mass of carbon dioxide ⇒ 44 grs/mol. Then,
44 grs ------------- 1 mol CO2
165 grs ------------ x = 3.75 moles of CO2.
Then, from Avogadros constant we have that there are 6.022 x 10^23 molecules in 1 mol. So,
1 mol CO2 ---------------- 6.022 x 10^23 molecules CO2
3.75 moles CO2-------- x = 2.258 x 10^24 molecules CO2.
There are 2.258 x 10^24 molecules CO2 in 165 grams of CO2.
If the Ka of HCN = 5.0 x 10^-10
Since
(Ka) (Ka) - 1 x 10^ -14
then
the Kb of its conjugate base (CN-) = 2.0 X 10^-5
since
pH + pOH = 14
when the pH = 10.00
then
the pOH = 4.00
& the OH-
would then equal 1.0 X 10^-4
NaCN as a base does a hydrolysis in water:
CN- & water --> HCN & OH-
notice that equal amounts of OH- & HCN are formed
Kb = [HCN] [OH-] / [CN-]
2.0 X 10^-5 = [1.0 X 10^-4] [1.0 X 10^-4] / [CN-]
[CN-] =(1.0 X 10^-8) / (2.0 X 10^-5)
[CN-] = (5.0 X 10^-4)
that's 0.00050 Molar
which is 0.00050 moles in each liter of aqueous KCN solution
which is
0.00025 moles KCN in 500. mL of aqueous KCN solution
use molar mass of KCN, to find grams:
(0.00025 moles KCN) (65.12 grams KCN / mole) = 0.01628 grams of KCn
which is 16.3 mg of KCN
& rounded to the 2 sig figs which are showing in the Ka of HCN , "5.0" X 10^-10
your answer would be
16 mg of KCN
sorry even after making a correction in calcs , I don't get one of your answers.
the only way that I could get one of them is to pretend that yours was a 1 sig fig problem,
in which case your 16 mg would round off to 20 mg.
but you have 3 sig figs in "500. ml", & 2 sig figs in both the "pH of 10.00."
& The Ka of HCN = "5.0 x 10^-10."
it does however take 12 mg of NaCN, to make 500. mL of aqueous solution pH of 10.00. the molar mass of NaCN has the smaller molar mass of 49.00 grams per mole.
maybe they meant NaCN, but wrote KCN instead.
I hope i answered this correctly for you.
I believe it’s B. At least thats what makes sense for me.
