The water used to water a golf course may be:
1) Carried off the surface of the golf course into nearby streams or rivers
2) Evaporated
3) Lost to the ground by the process of leaching
4) Absorbed by the vegetation in the golf course such as the grass, trees and shrubs.
Answer:
Cadmium +p= 48
Titanium p=22
Oxygen p=8
Krypton p=36
So the answer is Cadmium which has #48 protons
Explanation:
Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).
n = 0.250 mol; amount of substance.
V = 6.23 L; volume of the balloon.
T = 35°C = 308.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.
p = 1.015 atm; presure of the air.
Answer:
The technique is called weighing by difference
