At equilibrium the concentrations of:
[HSO₄⁻] = 0.10 M;
[SO₄²⁻] = 0.037 M;
[H⁺] = 0.037 M;
There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.
HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.
R
⇄ ![H^+ + SO_4^2^-](https://tex.z-dn.net/?f=H%5E%2B%20%2B%20SO_4%5E2%5E-)
I ![0.14](https://tex.z-dn.net/?f=0.14)
C
![+x](https://tex.z-dn.net/?f=%2Bx)
E
![x](https://tex.z-dn.net/?f=x)
×
for
. As a result,
![\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D.%20%5BSO_4%5E2%5E-%5D%7D%7BHSO_4%5E-%7D%20%3D%20K_a)
is large. It is no longer valid to approximate that
at equilibrium is the same as its initial value.
![\frac{x^2}{0.14-y} = 1.3 * 10^-^2](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%7D%7B0.14-y%7D%20%3D%201.3%20%2A%2010%5E-%5E2)
×
× ![10^-^2= 0](https://tex.z-dn.net/?f=10%5E-%5E2%3D%200)
Solving the quadratic equation for
since
represents a concentration;
![x=0.0366538](https://tex.z-dn.net/?f=x%3D0.0366538)
Then, round the results to 2 significant figure;
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Answer:
Empirical formula for the oxide is P₂O₃
Explanation:
Oxides from P, are:
P₂O; P₂O₃; P₂O₅; P₂O₇; P₄O₆; P₄O₁₀
The reaction is:
P₄ + 3O₂ → P₄O₆
Ratio between the white phosphorus and the oxide is 1:1
Molar mass P₄ = 123.89 g/m
Mass / Molar mass = mole
6.5 g/ 123.89 g/m = 0.0524 mole P₄
0.0524 mole of P₄O₆ have been formed.
Molar mass P₄O₆ = 219.89 g/m
Mol . molar mass = mass
0.0524 mole . 219.89 g/m = 11.52 g
Okay thanks for the update I will give you a call when you get home thanks
Answer:
=384.48°C
Explanation:
Change in enthalpy is negative because heat energy is lost to the surroundings.
Change in enthalpy= mass ×specific heat capacity× Change in temperature.
ΔH=MC∅
m=14grams
C of iron= 0.45 J/g°C
∅=?
186.0J= 14.0g × 0.45J/g°C×∅
186=6.3∅
∅=29.5°C
Final temperature= 414.0°C - 29.52°C
=384.48°C