Let the water evaporate and you'll be left with the sugar, or boil the water and collect the steam. the sugar will be left in the container you boiled the water in and you'll still have the water from the steam.
Answer:
a. 581.4 Pa
b. 3.33x10⁻⁴ mol/L
c. 3.49x10⁻⁴ mol/L
d. 0.015 g/L
Explanation:
a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:
pA = 0.9532*6.1
pA = 5.81452 mbar
pA = 5.814x10⁻³ bar
1 bar ----- 10000 Pa
5.814x10⁻³ bar--- pA
pA = 581.4 Pa
b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:
PV = nRT
Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.
n = PV/RT
n = (610*1)/(8.314*210)
n = 0.3494 mol
The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):
n = PV/RT
n = (581.4*0.9532)/(8.314*210)
n = 0.3174 mol
cA = n/V
cA = 0.3174/953.2
cA = 3.33x10⁻⁴ mol/L
c. c = ntotal/Vtotal
c = 0.3494/1000
c = 3.49x10⁻⁴ mol/L
d. The molar masses of the gases are:
CO₂: 44 g/mol
N₂: 28 g/mol
Ar: 40 g/mol
O₂: 32 g/mol
CO: 28 g/mol
The molar mass of the mixture is:
M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol
The mass concentration is the molar concentration multiplied by the molar mass:
3.49x10⁻⁴ mol/L * 43.36 g/mol
0.015 g/L
Answer:
See explanation
Explanation:
We must first write the equation of the reaction as follows;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Now;
We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles
So;
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
So;
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
We can now decide on the limiting reactant to be C3H8
Therefore;
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
Again;
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Hence;
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
In order to obtain the percentage yield from the reaction, we have;
b) Actual yield = 269.34 g
Theoretical yield = 396 g
Therefore;
% yield = actual yield/theoretical yield × 100/1
Substituting values
% yield = 269.34 g /396 g × 100
% yield = 68%
Molting
~~~hope this helps~~~
~~~davatar~~~
Answer:
B. 13.7 mm
Explanation:
by similarity of rectangular triangles, we have:
(1) Δ ABC and Δ AB'C'
∴ AB/AB' = AC/AC'
∴ hypotenuse (AB)² = AC² + BC².....Pythagorean theorem
⇒ AB² = (51 mm)² + (13 mm)²
⇒ AB² = 2770 mm²
⇒ AB = 52.63 mm
(2) Δ AB'C' :
∴ hypotenuse (AB')² = AC'² + B'C'²
∴ B'C' = 3.5 mm
from (1):
⇒ AB' = (AB)(AC') / AC = (52.63)(AC') / (51 mm)
⇒ AB' = 1.032AC'.......(3)
(3) in (2):
⇒ (1.032AC')² = AC'² + (3.5)²
⇒1.065AC'² = AC'² + 12.25
⇒ 0.065AC'² = 12.25
⇒ AC'² = 188.462
⇒ AC' = √188.462
⇒ AC' = 13.72 mm
