Answer:
have a great time at y avg be have a great time at all today I was going to answer restroom for spring break is over and over again and again and again and again and again and again your hhhha
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
chocolate
Explanation:
chocolate is 17,196 litres per 1kg.
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
= 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial 0.00525 0 0
change - 0.002625 +0.002625 +0.002625
equilibrium 0.002625 0.002625 0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
= -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
= 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
