Answer:
Total pressure 5.875 atm
Explanation:
The equation for above decomposition is
rate constant 
Half life 
Initial pressure 
Pressure after 3572 min = P
According to first order kinematics
solving for P we get
P = 2.35 atm
initial 4.70 0 0
change -2x +2x +x
final 4.70 -2x 2x x
pressure of
after first half life = 2.35 = 4.70 - 2x
x = 1.175
pressure of 
after first half life = 2x = 2(1.175) = 2.35 ATM
Total pressure = 2.35 + 2.35 + 1.175
= 5.875 atm
Answer:
T₁ = 39 K
Explanation:
Given data:
Initial pressure = 1023.6 kpa
Final pressure = 8114 kpa
Final temperature = 36°C (36+ 273= 309K)
Initial temperature = ?
Solution:
P₁/T₁ = P₂/T₂
T₁ = P₁×T₂ /P₂
T₁ = 1023.6 kpa × 309 K /8114 kpa
T₁ = 316292.4 K. Kpa /8114 kpa
T₁ = 39 K
Thus original pressure was 39 k.
3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
Answer:
a. 581.4 Pa
b. 3.33x10⁻⁴ mol/L
c. 3.49x10⁻⁴ mol/L
d. 0.015 g/L
Explanation:
a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:
pA = 0.9532*6.1
pA = 5.81452 mbar
pA = 5.814x10⁻³ bar
1 bar ----- 10000 Pa
5.814x10⁻³ bar--- pA
pA = 581.4 Pa
b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:
PV = nRT
Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.
n = PV/RT
n = (610*1)/(8.314*210)
n = 0.3494 mol
The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):
n = PV/RT
n = (581.4*0.9532)/(8.314*210)
n = 0.3174 mol
cA = n/V
cA = 0.3174/953.2
cA = 3.33x10⁻⁴ mol/L
c. c = ntotal/Vtotal
c = 0.3494/1000
c = 3.49x10⁻⁴ mol/L
d. The molar masses of the gases are:
CO₂: 44 g/mol
N₂: 28 g/mol
Ar: 40 g/mol
O₂: 32 g/mol
CO: 28 g/mol
The molar mass of the mixture is:
M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol
The mass concentration is the molar concentration multiplied by the molar mass:
3.49x10⁻⁴ mol/L * 43.36 g/mol
0.015 g/L
