Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
Answer:
0.00268 M
Explanation:
To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.
<u>Step 1:</u>
3.00 mL / 1,000 = 0.00300 L
Molarity = moles / volume (L)
0.0250 M = moles / 0.00300 L
(0.0250 M) x (0.00300 L) = moles
7.50 x 10⁻⁵ = moles
<u>Step 2:</u>
25.0 mL / 1,000 = 0.0250 L
0.0250 L + 0.00300 L = 0.0280 L
Molarity = moles / volume (L)
Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)
Molarity = 0.00268 M
Answer:The answer is C) The total mass of C and D only.
Answer:
65.4%
Explanation:
The redox reaction is a 1:1:1 reaction because the reagents suffer a double displacement reaction, and the substance that is substituted have the same charge (H+ and Br-), thus, we first need to know which of the reagents is the limiting.
Let's test the 4-nitrobenzaldehyde as the limiting. The mass needed for sodium borohydride (m) is the mass given of 4-nitrobenzaldehyde multiplied by the stoichiometric mass of sodium borohydride divided by the stoichiometric mass of 4-nitrobenzaldehyde. The stoichiometric mass is the number of moles in the stoichiometric representation (1:1:1) multiplied by the molar mass, so:
m = (4.13 * 37.83*1)/(151.12*1)
m = 1.034 g
So, the mass needed of the other reagent is larger than the mass that was given, so, it will be the limiting, and the stoichiometric calculus must be done with it.
The mass of the product that was expected is then:
m = (0.700*153.14*1)/(37.83*1)
m = 2.83 g
The percent yield is the mass that was formed divided by the expected mass, and then multiplied by 100%:
%yield = (1.85/2.83)*100%
%yield = 65.4%