<h2>Answer:</h2>
B) endothermic reaction.
<h2>Explanation:</h2>
Melting of ice is endothermic because it is taking heat from the environment, feeling cold, because it requires energy to break the ice bonds. It's also why you feel cold when wet it takes energy to evaporate water. The ice absorbs the energy from the environment. Its internal potential energy increases, therefore, it's endothermic. It also increases the entropy of reaction.
Answer:
d. K<1 E∘cell is negative
Explanation:
Since E⁰ = negative , ΔG = -nFE⁰ = -nF -ve = +ve.
Also, ΔG = -RTlnK
K = exp(-RTΔG)
Since ΔG = +ve, -RTΔG = -ve
K = 1/exp(RTΔG) < 1.
So our answer is E⁰ cell is negative and K < 1
Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = 
Also, when number of moles are equal in a solution then the formula will be as follows.

It is given that
is 8.00 M,
is 7.00 mL, and
is 0.80 M.
Hence, calculate the value of
using above formula as follows.



= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
Answer:
See explanation below
Explanation:
The first two pictures show the reagents used in these reactions a) and b). As it was stated, An E2 reaction proceeds with an antiperiplanar stereochemistry, so in the case of reaction a) it fill form a product with the groups in opposite directions. In other words, a Trans product.
In the case of reaction b) we have the same reaction, with the difference that we have changed the CH3 and phenyl group of positions. This will cause that the reaction will proceed the same but the stereochemistry of the final product will be changed too. In this case, and according to the picture 3 attached, we can see that the product formed is a cis product. So we can conclude that the relation of product a) and b) is that they are isomers, the trans and cis isomers respectively. See picture below for mechanism and products
Answer:
D. Anti-periplanar
Explanation:
In the <u>second step</u> of the intramolecular William Ether Synthesis mechanism (figure 1) we will have the attack of the negative charge of the oxygen to the carbon bond to the Br. At the same time the Br leaves, so a bond would be broken (the <u>C-Br</u> bond) and a bond would be formed (the <u>C-O</u> bond).
Now, this process can happen only if the <u>attack</u> and the <u>leaving group </u>has an anti configuration (figure 2). In an anti configuration the <u>nucleophile</u> and the <u>leaving group</u> would have <u>opposite directions</u>.