Answer:
E) A, B, and C
Explanation:
Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.
Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.
Answer:
D. is the Answer. ✅
Explanation:
The Moon is 1/4 the size of Earth.
A is not right because Moon isn't a dwarf planet.❌
B is not right because the Moon isn't 1/2 the size of Earth.❌
C is not right because the Moon is no where near as large as Earth.❌
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)
Answer:
All right. So let's calculate the density of a glass marble. Remember that the formula for density is mass over volume. So if I know that the masses 18.5 g. And I know that the um volume is 6.45 cubic centimeters. I can go ahead and answer this to three significant figures. So it's going to be 2.87 grams per cubic centimeter. Okay, that's our density. Now, density is an intensive process. Okay. We're an intensive property. I really should say. It doesn't depend on how much you have. Mhm. If I have one marble, its density is going to be 2.87 g per cubic centimeter. If I have two marbles, the density will be the same because I'll double the mass and I'll also double the volume. So when I divide them I'll get the same number. Okay, that's what makes it an intensive property. No matter how many marbles I have, they'll have the same density. Mass though is not an intensive property. So if I have six marbles and I want to know what the massive six marbles is. Well, I know the mass of each marble is 18.5 g. So the mass of six marbles Is going to be 100 11 g. Because mass is an extensive property. It depends on how much you have. If I change the number of marbles, I'm going to change the mass. That's an extensive property. All right. So we've calculated the density. We've calculated the mass and then what happens to the density of one marble compared to six marbles as we mentioned before. Since densities and intensive property, the densities will be the same, no matter how may.
Explanation: