Answer:
28atm
Explanation:
Using Gay lussac's law equation as follows:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
T1 = initial temperature (K)
P2 = final pressure (atm)
T2 = final temperature (K)
Based on the information provided in this question;
P1 = 30.0 atm, T1 = 30.0°C, P2 = ?, T2 = 10.0°C
NOTE: Absolute temperature i.e. Kelvin is required for this law
T1 = 30°C + 273K = 303K
T2 = 10°C + 273K = 283K
Using P1/T1 = P2/T2
30/303 = P2/283
Cross multiply
P2 × 303 = 30 × 283
303P2 = 8490
P2 = 8490/303
P2 = 28.02
New pressure of the gas = 28atm
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
C3H8 + 5O2 ------> 3CO2 + 4H2O
from reaction 1 mol 5 mol
given 1.82 mol x mol
x=(1.82*5)/1 = 9.10 mol CO2
Answer:
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Explanation:
