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Brums [2.3K]
2 years ago
11

How much volume (in cm3) is gained by a person who gains 12.3 lb of pure fat?

Chemistry
1 answer:
maks197457 [2]2 years ago
4 0

Answer:

So we are given with the mass while we are asked for the amount of volume. So this means, we must need an information on the density. From literature, the density of human fat is 0.918 g/cm³. Convert grams to lb by the conversion that 1,000 g = 2.2 lbs.

Density = 0.918 g/cm³ * (2.2 lbs/1,000 g) = 0.0020196 lb/cm³

Volume = Mass/Density = 12.2 lb / 0.0020196 lb/cm³

Volume = 6,040.8 cm³

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When solid ammonium chloride dissociates at a certain temperature in a 0.500 dm3 container, ammonia and hydrogen chloride are fo
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1) Reaction

<span>NH4Cl(s) ---> NH3(g) + HCl(g)


2) equilibrium equation, Kc


Kc = [NH3] * [HCl]


3) Table of equilibrium formation



step               concentrations
                       </span>
<span>                            NH4Cl(s)                     NH3(g)             HCl(g)


start                      1.000 mole                     0                    0


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produce                                                       +x                  + x
                         ------------------                ----------              -----------

end                         1 - x                             +x                    +x
            

1 - x = 0.3 => x = 1 - 0.3 = 0.7


[NH3] = [HCl] = 0.7/0.5 liter = 1.4                (I used 0.500 dm^3 = 0.5 liter)


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Kc = [NH3] [HCl] = (1.4)^2  = 1.96


Which is the number that you were looking for.


Answer: Kc = 1.96
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4 years ago
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam
nikklg [1K]
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)
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