A) a good hypothesis to be put in question for the experiment would be “Fluorescent light can fade color photographs quicker than direct sunlight can.”
B) the controlled variables, or things that will need to be kept constant throughout the experiment, would be the amount of light on each picture, the amount of ink on each picture, the type of ink on each picture, the size of each picture, and the time of day of the light for each picture.
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
NH4NO3 b) C8H8O4 c) O3 d) C3H5(NO3)3
Per hour-68x60=4080
Per day-4080x24=97920
Per week-97920x7=685440
9.0x10^5 = 900,000
850,000 - 900,000 = -50,000
-50,000 = -5.0x10^4 = Answer
