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Goshia [24]
3 years ago
10

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st

ems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0269 m solution of oxalic acid?
Chemistry
1 answer:
VLD [36.1K]3 years ago
3 0
Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.

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Explanation:

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The balanced equation will be:

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8 0
3 years ago
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Answer:

a) the final kilocalories per gram for food will be less because the mass was reduced

b)the final kilocalories per gram for food will be less since

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d) the final kilocalories per gram for food will be more.

Explanation:

a) the final kilocalories per gram for food will be less because the mass was reduced from 110.3 to 101.3g

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3 years ago
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Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of water = 500.0 g

c = specific heat of water = 4.18J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 37.0^oC

Now put all the given value in the above formula, we get:

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

Q=25080J=25.1kJ

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Explanation:

8 0
2 years ago
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