Answer:1 3 4 1 im pretty positive
Solution:
At the equivalence point, moles NaOH = moles benzoic acid
HA + NaOH ==> NaA + H2O where HA is benzoic acid
At the equivalence point, all the benzoic acid ==> sodium benzoate
A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)
Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11
Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150
x^2 = 3.33x10^-12
x = 1.8x10^-6 = [OH-]
pOH = -log [OH-] = 5.74
pH = 14 - pOH = 8.26
The following is produced when propane (C₃H₈) is combusted completely : H₂O
<h3>Further explanation </h3>
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (especially alkanes) 
For combustion of propane C₃H₈ (n = 3) ⇒ completely(excess O₂) :
C₃H₈+5O₂⇒3CO₂+4H₂O
The products of combustion : CO₂ and H₂O
Answer:
2.7 x 10^-19 J
Explanation:
The formula needed for this problem is
E = hν
where E = energy, h = Planck's constant = 6.626x10^-34 and ν is the frequency
c = λν
where c = speed of light = 3x10^8, and λ = wavelength
3x10^8 = 7.35x10^-7 . ν
ν = 4.08 x 10^14 Hz
E = 6.626x10^-34 . 4.08x10^14 = 2.7 x 10^-19 J