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By using ICE table:
CH3NH3+ + H2O → CH3NH4 2+ + OH-
initial 0.175 0 0
change -X +X +X
Equ (0.175-X) X X
when: Ka = Kw / Kb
= (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11
when Ka = [CH3NH42+][OH-] / [CH3NH3+]
by substitution:
2.3 x 10^-11 = X^2 / (0.175 - X ) by solving for X
∴ X = 2 x 10^-6
∴[OH-] = 2 x 10^-6
∴POH = -㏒[OH-]
= -㏒(2 x 10^-6)
= 5.7
when PH + POH = 14
∴PH = 14 - 5.7 = 8.3
Answer: 116 calories of energy
Explanation:
A calorie is a non-standard unit of energy.
On combustion,
1 gram Carbohydrates = 4 calories,
1 gram protein = 4 calories,
1gram fat = 9 calories.
Therefore,
15 grams of carbohydrates = (15*4) = 60
5 grams of protein = (5*4) = 20
4 grams of fats = (4*9) = 36
Then add up: 60 + 20 + 36 = 116 calories of energy
Answer:
Phase C - Liquid State
Phase E - Gaseous State
Explanation:
Usually, in phases of water, we have the following;
When temperature is less than zero, it is said to be in its solid phase as ice.
When temperature is between 0 to 100, we can say it is in the liquid phase as water.
When temperature is above 100°C, It is said to be in the gaseous phase as vapour.
From the diagram;
Phase C is the only liquid state because it falls between temperature of 0°C and 100°
Also, only phase E is in the gaseous phase because the temperature is above 100°C.
Answer:
Option e and f are possible
Explanation:
Since we know that 0.264 gallon = 1L ⇒ we change this in all equations.
8.08 L * A =
a) 8.08 L * 0.264 gal / 8.08 L = 0.264 gal = 1L
b) 8.08 L * 8.08 gallon / 0.264L = (8,08/ 0.264) L/0.264 L = 30.61 / 0.264 = 115.93 L *8.08 L
c) 8.08 L * 8.08 gallon / 1L = (8,08/ 0.264) L/1 L = 30.61 L * 8.08 L
d) 8.08 L * 0.264 L / 1 gallon =8.08 L * 0.264 L / 1 gal
e) 8.08 L * 0.264 gallon / 1L = 8.08 L *1L / 1L = 8.08 L
f) 8.08 L * 1L / 0.264 gallon = 8.08L * 1L / 1L = 8.08 L
The last 2 options are possible ( e and f )
