Answer: 59.24 atm
Explanation:
Given that:
Original Volume of gas V1 = 2.7L
Temperature T1 = 42.7°C
Convert Celsius to Kelvin
(42.7°C + 273 = 315.7K)
Pressure P1 = 684.9 torr
Final Volume V2 = 0.14 L
Final temperature T2 = 803.1°C
Convert Celsius to Kelvin
(803.1°C + 273 = 1076.1K)
Final pressure = ?
Then, apply the combined gas equation
(P1V1)/T1 = (P2V2)/T2
(684.9 torr x 2.7L)/315.7K = (P2 x 0.14L)/1076.1K
1849.23/315.7 = 0.14P2/1076.1
Then, cross multiply
1849.23 x 1076.1 = 315.7 x 0.14P2
1989956.403 = 44.198P2
Divide both sides by 44.198
1989956.403/44.198 = 44.198P2/44.198
45023.67 torr = P2
Now, convert the pressure in torr to atmosphere
If 760 torr = 1 atm
45023.67 torr = Z atm
Then, cross multiply
760 torr x Z = 45023.67 torr x 1 atm
Z = 45023.67 torr / 760 torr
Z = 59.24 atm
Thus, the new pressure of the gas will be 59.24 atm
Answer:
112.92 s.
Explanation:
Let M₁ be the molar mass of N₂O
Let t₁ be the time taken for N₂O to effuse.
Let M₂ be the molar mass of I₂
Let t₂ be the time taken for I₂ to effuse.
Molar mass (M₁) of N₂O = (14×2) + 16 = 28 + 16 = 44 g/mol
Time (t₁) of effusion of N₂O = 47 s
Molar mass (M₂) of I₂ = 127 × 2 = 254 g/mol
Time (t₂ ) of effusion of I₂ =?
The time take for the same amount of I₂ to effuse can be obtained as follow:
t₂/t₁ = √(M₂/M₁)
t₂/47 = √(254 / 44)
Cross multiply
t₂ = 47 × √(254 / 44)
t₂ = 112.92 s
Therefore, it will take 112.92 s for the same amount of I₂ to effuse.
Answer:
Reduction half
Fe2+ +2e -------> Fe
Oxidation half
Mn-----> Mn4+. +4e
Explanation:
Oxidation is loss of electron and reduction is gain of electron
Answer:
i am very sorry i have no answer but if i get one i can tell u the answer
Explanation:
C funnel because the funnel would have a large enough entrance to put a solid through
