Answer:
104.352°C
Explanation:
Data Given:
Boiling point of water = 100.0°C
Kb (boiling point constant = 0.512°C/m
Concentration of the Mg₃(PO₄)₂ = 8.5 m
Solution:
Formula Used to find out boiling point
ΔTb = m.Kb . . . . . . (1)
where
ΔTb = boiling point of solution - boiling point of water
So,
we can write equation 1 as under
ΔTb = Tb (Solution) -Tb (water)
As we have to find out boiling point so rearrange the above equation
Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)
Put values in Equation 2
Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C
Tb (Solution) = 4.352 + 100.0°C
Tb (Solution) = 104.352°C
so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C
Answer:
Explanation:
MM: 2.016 17.03
N₂ + 3H₂ ⟶ 2NH3
m/g: 26.3
1. Theoretical yield
(a) Moles of H₂
(b) Moles of NH₃
(c) Theoretical yield of NH₃
(d) Percent yield
Answer:
1. 0.97 V
2.
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
Remark
The question with these kind of problems is "Which R do you use?" That's where dimensional analysis is so handy. You must look at the units of the givens and choose your R accordingly. You'll see how that works in a moment.
You need to list the givens along with their units and in this case the property you want to solve for. You need all that to determine the R value
Givens
n = 0.25 moles
T = 35°C = 35 + 273.15 = 308.15°K
V = 6.23 L
Pressure = P in kPa
Which R
The units of the R you want has to have units of moles, kPa, °K and liters
The R that you want is 8.314
<em><u>Formula</u></em>
PV = nRT
P 6.23 = 0.25 * 8.314 * 308.15 Combine the left
P*6.23 = 640.5
P = 640.5/6.23 = 102.81 The answer should be 100 kpA of 1.0 * 10^2 kPa
because the number of moles has only 2 sig digs.
But if sig digs are not a problem 102.8 is likely close enough.
Second Question
You are going to have to clean up the numbers. I think I've got only 1 chance at this. The partial pressures of the 2 gases will add up to the total pressure. So the total pressure was 100 approx and the water vapor was 3.36 kPa. The difference is
Total = air + water vapor
100.18 = air + 3.36 about Subtract 3.36 from both sides.
100.18 - 3.36 = 96.82 about. Pick the answer that is closest to that. I'll clean up the numbers if I can.
Answer C
There can be three definitions of a base: Arrhenius base, Lewis base or Bronsted-Lowry base. Isoquinoline acts specifically as a Bronsted-Lowry base, which is a proton acceptor. Also, it acts as an Arrhenius base because it produces OH- ions after the reaction. The net ionic equation is:
<span><em>C</em></span><em>₉</em><span><em>H</em></span><em>₇</em><span><em>N (aq) + H</em></span><em>₂</em><span><em>O ---> C</em></span><em>₉</em><span><em>H</em></span><em>₇</em><span><em>NH</em></span><em>⁺</em><span><em> (aq) + OH</em></span><em>⁻</em><span><em> (aq)</em></span>