Answer:
Explanation: Bromine, the dark red color disappears quickly as the atoms of bromine bond with the atoms of carbon in the double bond.
We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:
Kw = Kb * Ka
where Kw is always equal to 1x10^-14
so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:
1x10^-14 = 1.4x10^-9 * Ka
solve for Ka
Ka = 7.14x10^-6
Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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Data:
V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm
V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?
Formula:
Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2
=> P2 = P1 V1 T2 / (T1 V2)
P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)
P2 = 3.22 atm
Answer:
33.95 grams of NaN3
Explanation:
Number of moles of NaN3 = mass (m)/MW = m/65 mole
I mole of NaN3 requires 22.4L air bag
m/65 mole of NaN3 required 11.7L
22.4m/65 = 11.7
22.4m = 65×11.7
22.4m = 760.5
m = 760.5/22.4 = 33.95grams of NaN3