Answer:
6 sig figs
Explanation:
sig figs are numbers that are more than 0, but in this case, after the decimal, the 0's at the end count as sig figs as well
The answer is the atomic mass.
Answer:
Methanol is more acidic than the alkyne and will be deprotonated instead.
Explanation:
of methanol is around 15 and of terminal alkyne is around 26.
, where is acid dissociation constant
So, higher the acidity of an acid, higher will its value and thereby lower will be its value.
So, methanol is certainly stronger acid than terminal alkyne.
Hence sodium amide preferably deprotonates methanol instead of terminal alkyne.
Hence, option (A) is correct.
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
= 600 gallons
So
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
= 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
Answer:
See Explanation
Explanation:
a. pH of 1M HOAc(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 1.0M x x
Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵
=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M
=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37
b. pH of 0.10M CH₃NH₃OH(aq)
CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴
C(eq) 0.10M x x
=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M
=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M
=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18
=> pH = 14 – pOH = 14 – 2.18 = 11.82
c. pH of 0.30M HOAc/0.10M OAcˉ(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 0.30M x 0.10M
=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]
= 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M
=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26