Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
1. No
2.No
I hope this helps:)
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹
How many moles of water at produced when 5 moles of hydrogen gas react with 2 moles of oxygen gas the answer is D
Answer:
what the acctual fhuck typa chinese algorithim is dis, youve got me bamboozled frffr
Explanation:
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__5___SiO2 + ___2___CaC2 → __5___Si + ____2___CaO + __4___CO2
