We can calculate the final temperature from this formula :
when Tf = (V1* T1) +(V2* T2) / (V1+ V2)
when V1 is the first volume of water = 5 L
and V2 is the second volume of water = 60 L
and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K
and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K
and Tf is the final temperature of water in Kelvin
so, by substitution:
Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)
= 1765 + 18180 / 65 L
= 306 K
= 306 -273 = 33° C
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
I would agree with the second one, not the first. You can't always see the chemical reaction, and it isn't always sudden. But the second claim is true.
Answer:
sorry I have no idea what the answer is
Answer:
151 g/mol
Explanation:
In order to solve this problem we need to keep in mind the formula for the <em>boiling point elevation</em>:
Where:
- ΔT is the temperature difference between the boiling point of the solution and that of pure water. 100.37 °C - 100.00 °C = 0.37 °C.
- <em>m</em> is the molarity of the solution
- i is the van't Hoff factor. As the solute is a nonelectrolyte, the factor is 1.
Input the data and <u>calculate </u><em><u>m</u></em>:
- 0.37 °C = 0.51 °C/m * <em>m</em> * 1
We now can <u>calculate the number of moles of the substance</u>, using the <em>definition of molarity</em>:
- molarity = moles of solute / kg of solvent
In this case kg of solvent = 90.0 g / 1000 = 0.090 kg
- 0.72 m = moles / 0.090 kg
Finally we <u>calculate the molar mass</u>, using the <em>number of moles and the mass</em>:
- 9.81 g / 0.065 mol = 151 g/mol
