<span>Answer:
1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>
They are different by a phase shift of pi/2
Find out how much the donuts cost then divide that by $10,000 and you will get your answer
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
The hydrocarbon is used in excess.
<h3><u>Explanation</u>:</h3>
The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.
