Answer:
Explanation:
If the enzyme active site is complementary to the substrate conformation rather than to the transition state, it is unlikely that the reaction will proceed and release a product, because the enzyme-substrate complex will be tightly bound (ΔG will raise).
On the other hand, when the enzyme active site is complementary to the transition state, the substrate will not be tightly bound and will be more prone to be transformed into the product (<u>ΔG will be lowered</u>) and afterward, be released.
The weak interactions (non-covalent bonds) will stabilize the energy of the transition state and reduce its energy, thus lowering the activation energy). If the transition state is stable, it will form more easily and<u> the reaction will be more likely to proceed.</u>
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Answer:
Here's what I get
Explanation:
A. Initial observation
Gary's shell had slime and an odour.
B. Independent variable
The independent variable is the one that the experimenter changes.
There are two independent variables: the rubbing with seaweed and the drinking of Dr. Kelp.
C. The dependent variable
The dependent variable is the amount of slime and odour.
D. The conclusion
Sponge Bob can conclude that rubbing the shell with seaweed and drinking Dr. Kelp removes the slime and odour.
However, this was a poorly designed experiment. He doesn't know if it is the seaweed or the Dr. Kelp that gives the result or if he must use both together. He should change only one independent variable at a time.
Answer:
16.56 g
Explanation:
Mass is the production of Volume and Density.
m = V. d = 6 × 2.76 = 16.56 g
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
