mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answer:
C ≈ 1.44 × 10⁻³ mol/m³
Explanation:
The given information are;
The liquid volume the pond can hold = 104 m³
The volume of inflow into the pond = 5 m³/h
The volume of outflow into the pond = 5 m³/h
The concentration of the chemical in the inflow water = 0.01 mol/m³
The concentration of the chemical discharged directly into the water = 0.1 mol/h
The concentration, 
, of chemical that enters the water through inflow per hour is given as follows;
= 0.01 mol/m³ × 5 m³/h = 0.05 mol/h
The concentration, 
, of chemical that enters the water through direct discharge per hour is given as follows;
= 0.1 mol/h
The total concentration that enters the pond per hour is given as follows;
+ = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h
Whereby the water in the pond properly mixes with the pond, we have;
The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³
C ≈ 1.44 × 10⁻³ mol/m³.
Answer: 64.6 mmHg
Explanation:
Given that:
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
Thus, the pressure of the gas is 64.6 mm hg
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The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.
