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Answer:
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Explanation:
for the reaction
PCl₅(g) → PCl₃(g) + Cl₂(g)
where
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M
and [A] denote concentrations of A
if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M
therefore
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M
[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M
[PCl₅] = 3.64*10⁻³ M
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Answer:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
