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 
⇄ 
I 
C 
E 
× 
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 It is no longer valid to approximate that ![[HSO^-_4]](https://tex.z-dn.net/?f=%5BHSO%5E-_4%5D)
at equilibrium is the same as its initial value.
× 
× 
Solving the quadratic equation for 
since represents a concentration;
Then, round the results to 2 significant figure;
Learn more about concentration here:
brainly.com/question/14469428
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Answer:
The answer to your question is m = 4.7 kg
Explanation:
Data
Ice Water
mass = ? mass = 711 g
T₁ = -13°C T₁ = 87°C
T₂ = 10°C T₂ = 10°C
Ch = 2090 J/kg°K Cw = 4180 J/kg°K
Process
1.- Convert temperature to kelvin
T₁ = 273 + (-13) = 260°K
T₁ water = 87 + 273 = 360 °K
T₂ = 10 + 273 = 283°K
2.- Write the equation of interchange of heat
- Heat lost = Heat absorbed
- mwCw(T₂ - T₁) = miCi(T₂ - T₁)
-Substitution
- 0.711(4180)(10 - 87) = m(2090)(10 - (-13))
- Simplification
228842.46 = 48070m
m = 228842.46/48070
-Result
m = 4.7 kg
Answer:
0.500 mol/dm³
Explanation:
Using the formula below;
CaVa = CbVb
Where;
Ca = concentration of acid (mol/dm³)
Cb = concentration of base (mol/dm³)
Va = volume of acid (cm³)
Vb = volume of base (cm³)
In accordance to the information provided in this question is;
Va = 5cm³
Vb = 250 cm³
Ca = 12 mol/dm³
Cb = ?
Using CaVa = CbVb
12 × 5 = Cb × 250
60 = 120Cb
Cb = 60/120
Cb = 0.500 mol/dm³
Hope this helps! If you dont understand balancing equations in general, say so in the comments, I’m happy to help
