Answer:
Explanation: the module will end up working out when you mix NO2CI and CI
Answer:
60 grams
Explanation:
We have the balanced equation (without state symbols):
6
H
2
O
+
6
C
O
2
→
C
6
H
12
O
6
+
6
O
2
So, we would need six moles of carbon dioxide to fully produce one mole of glucose.
Here, we got
88
g
of carbon dioxide, and we need to convert it into moles.
Carbon dioxide has a molar mass of
44
g/mol
. So here, there exist
88
g
44
g
/mol
=
2
mol
Since there are two moles of
C
O
2
, we can produce
2
6
⋅
1
=
1
3
moles of glucose
(
C
6
H
12
O
6
)
.
We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.
Glucose has a molar mass of
180.156
g/mol
. So here, the mass of glucose produced is
1
3
mol
⋅
180.156
g
mol
≈
60
g
to the nearest whole number.
So, approximately
60
grams of glucose will be produced.
PH= pKa + log( conjugate base/ conjugate acid)
Reaction of
Acyl Halides with
Grignard reagent results in the formation of
Ketones in first step. While in second step reaction of Grignard reagent with Ketones results in the formation of
Tertiary Alcohols.
If you want to
stop the reaction at
Ketone stage then you are required to use another
mild reactive organometallic compound. In our case we will use Organocuprates.
Organocuprates are also known as Gilman Reagents. These reagents does not add to ketones, aldehydes and esters but they can add to acid halides to produce Ketones.