Answer:
When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.
Aldolase cleaves a hexose into two trioses.
[See the image attached].
Asterisk indicates the label.
When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.
Answer:
look at the graph
Explanation:
We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.
Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.
So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)
Answer:
Rise surface fall surface rise
Explanation:
Hope this helps!
Answer:
ΔE = -2661 KJ/mole
ΔH = -2658 KJ/mole
Explanation:
ΔH = q - PΔV
ΔE = q + w
<u>First, to find ΔE:</u>
The reaction PRODUCES 2658 kJ of h (q), and does 3 kJ of work (w).
2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction <u><em>PRODUCES</em></u> heat, which means ΔE is negative.
ΔE = -2661 KJ/mole
<u>Second, to find ΔH:</u>
ΔH = q - PΔV
ΔH = 2658 kJ(q) - PΔV
Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.
ΔH = 2658 KJ(q) - (0)ΔV
ΔH = 2658 KJ - 0
ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.
ΔH = -2658 KJ/mole
I hope this helped! Have a nice week.
Answer:False
Explanation:
slow down/ meet and obstacle.