When Wolverine’s 10-pound adamantium claws are dissolved in 100 mL of 10 M nitric acid, 10.7 grams of adamantium nitrate are rec
1 answer:
The percent yield is 71.3 %.
Explanation:
Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.
So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.
Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.
Thus, the percent yield is 71.3 %.
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Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-
where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas =
R = Gas constant =
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:
According to the reaction:-
1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of = 0.01017 moles
Molar mass of = 86.93685 g/mol
So,
Applying values, we get that:-
<u>0.88 g of should be added to excess HCl (aq) to obtain 235 mL of
at 25 degrees C and 805 Torr.</u>
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
.
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:
.
---
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.