Answer:
C
Explanation:
All thing in option c is flammable
Answer:
0.78 atm
Explanation:
Step 1:
Data obtained from the question. This includes:
Mass of CO2 = 5.6g
Volume (V) = 4L
Temperature (T) =300K
Pressure (P) =?
Step 2:
Determination of the number of mole of CO2.
This is illustrated below:
Mass of CO2 = 5.6g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Number of mole CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 5.6/44
Number of mole of CO2 = 0.127 mole
Step 3:
Determination of the pressure in the container.
The pressure in the container can be obtained by applying the ideal gas equation as follow:
PV = nRT
The gas constant (R) = 0.082atm.L/Kmol
The number of mole (n) = 0.127 mole
P x 4 = 0.127 x 0.082 x 300
Divide both side by 4
P = (0.127 x 0.082 x 300) /4
P = 0.78 atm
Therefore, the pressure in the container is
Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.
Answer:
The answer to your question is: T2 = 235.44 °K
Explanation:
Data
V1 = 3.15 L V2 = 2.78 L
P1 = 2.40 atm P2 = 1.97 atm
T1 = 325°K T2 = ?
Formula
Process
T2 = (P2V2T1) / (P1V1)
T2 = (1.97x 2.78x 325) / (2.40 x 3.15)
T2 = 1779.895 / 7.56
T2 = 235.44 °K
T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
