Answer: Well!
Explanation: I was going to answer D but fverdell82156 got to it first! So I have to agree with him! It is D!
What do you mean by that Oh wait do you mean like 2×2=4 or 2+2=4 something like that maby?? I am still confused?? XD
1) <u>Stereo-selective (or enantioselective)</u> reactions form predominately or exclusively one enantiomer.
2) Epoxidation is the addition of a single oxygen atom to an alkene to form an epoxide.
3) <u>Hydrogenation (or reduction)</u> of an alkene forms an alkane by addition of H₂.
4) <u>Dihydroxylation</u> is the addition of two hydroxy groups to a double forming, a 1,2-diol or glycol.
5) <u>oxidative</u> cleavage of an alkene breaks both the σ and π bonds of the double bond to form two carbonyl groups.
6) <u>Regioselective</u> reactions form predominately or exclusively one constitutional isomer.
7) <u>Syn</u> dihydroxylation results when an alkene is treated KMnO4 or OsO4, where each reagent adds two oxygen atoms to the same side of the double bond.
Answer
A. It changes the rate, R
Explanation
When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol