The pH of pure water is 7.0, which is neutral.
For a pH below this, the water is acidic. Substances that are acidic are often corrosive and thus could cause weather damaging. The pH of the precipitation in NY is below that of the precipitation in IL, and NY receives more precipitation, so for both of those reasons, it is likely to have more chemical weathering.
The heat of solution is -51.8 kJ/mol
<h3>What is the heat of solution?</h3>
We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.
Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles
Temperature rise = 26.0 ∘c
Mass of the water = 100.0 grams
Heat capacity = 4.184 j/g⋅°c
Then;
ΔH = mcθ
ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ
Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol
Learn more about heat of solution:brainly.com/question/24243878
#SPJ1
The answer is 4.69 x 10⁻¹⁹ I hope this helped!
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
Answer:
pKa of the acid HA with given equilibrium concentrations is 6.8
Explanation:
The dissolution reaction is:
HA ⇔ H⁺ + A⁻
So at equilibrium, Ka is calculated as below
Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260
= 15.38 x 10⁻⁸
Hence, by definition,
pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813
