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: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
Answer:
Cyclopropane has a planar carbon back bone while propane does not
Explanation:
We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.
However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.
Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
