Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Answer:
m = 32.34 pounds of ice.
Explanation:
In this case we need to use the following expression of heat:
q = m * ΔHf (1)
Where:
q: heat absorbed in J or kJ
m: mass of the compound in g
ΔHf: heat of fusion of the water in kJ/g
We are asked to look for the mass of ice in pounds, so after we get the grams, we need to convert the grams to pounds, using the following conversion:
1 pound --------> 453.59 g (2)
So, we have the heat and heat of fusion, from (1) let's solve for the mass, and then, using (2) the conversion to pounds:
q = m * ΔHf
m = q / ΔHf
m = 4900 / 0.334 = 14,670.66 g of ice
Now, the conversion to pounds:
m = 14,670.66 g * 1 pound/453.59 g
<h2>
m = 32.34 pounds of ice.</h2>
Hope this helps
HCl = 1 + 35.5 => 36.5 g/mol
hope this helps!
Idk if this is the information you need but the energy level gets higher the farther you move down the Periodic Table. Every element in a vertical line has the same energy level.
