Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
Answer:

Explanation:
In this case, we have to start with the <u>chemical reaction</u>:

So, if we start with <u>10 mol of cyclohexanol</u> (
) we will obtain 10 mol of cyclohexanol (
). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

With this value we can calculate the grams:

Now, we have as a product 500 mL of
. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

Finally, with these values we can calculate the <u>yield</u>:
%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
A B and C are all chemical changes. paper tearing is not.
First, you need to convert kg to g.
So, 1 kg =1000g.
3.5 x 1000 = 3500g Ca(OH)2
We need to know the molar mass of Ca(OH)2.
Ca= 40.08 g
O=2(15.999)
H=2(1.0079)
Add them all together and you get 74.0938 g.
Put it in the formula from mass to moles.
# of moles = grams Ca(OH)2 x 1 mol Ca(OH)2
--------------------
molar mass Ca(OH)2
3500 g Ca(OH)2 x 1 mol Ca(OH)2
---------------------
74.0938 g Ca(OH)2
So divide 1/74.0938 and multiply by 3500.
You will get about 47.24 moles Ca(OH)2.
Hope this helps! :)
Answer:
If a body is accelerated by 10m/s^2, it means that the body's speed (or velocity to be precise) is increasing by 10m/s every second. That is, if the initial ...