i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
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ii. Given the dissolution of some substance
,
the Ksp, or the solubility product constant, of the preceding equation takes the general form
![K_{sp} = [B]^y [C]^z](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BB%5D%5Ey%20%5BC%5D%5Ez)
.
The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.
So, given our dissociation equation in question i., our Ksp expression would be written as:
![K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Cmathrm%7B%5BPb%5E%7B2%2B%7D%5D%20%5BSO_4%5E%7B2-%7D%5D%7D)
.
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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).
We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:
So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.
A prediction is not based upon evidence or reliable tests. A prediction is a guess as to what will happen in the future. Forecasts for example are nothing more than prediction, as even with "meteorological evidence", nothing is ever truly sure when in regard to wheather.
So "A"
Answer:
See explanation
Explanation:
First, we need to use the correct expression:
Q = m*Cp*ΔT (1)
Q = n*ΔH (2)
These are the 2 expressions to calculate heat or energy.
Now, we want to know the change of temperature of the nitrate in water after being added, so with the innitial data of nitrate, we can calculate heat using the second expression. First, we need to calculate moles with the molecular mass:
n = m/MM
n = 42/80.1 = 0.52 moles
With these moles, we can calculate heat with the ΔH of this reaction:
Q = 0.52 * 25.7 = 13.364 kJ or 13,364 J
However, this heat as is being absorbed, the value would be negative.
Now that we have heat, we can use expression (1) and plug these values to solve for ΔT, but before, we need to know the total mass of the solution (water + nitrate)
m = 250 + 42 = 292 g
now, solving for ΔT:
-13,364 = 292 * 4.18 * ΔT
ΔT = -13,363 / (292 * 4.18)
ΔT = -10.95 °C
HNO3 is a strong acid and NaOH is a
strong base. An equal number of moles of
each will create a neutral solution.
(.1500L)(.100M NaOH) = .01500 moles
NaOH, and therefore .015 moles OH-
.01500 moles H+ = .01500 moles HNO3 =
(.120M HNO3)(volume HNO3)
volume HNO3 = .125L HNO3, or 125mL
HNO3
