If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most
1 answer:
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Answer:
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Given: Kc = 0.140
Temperature = 1778 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (1778 + 273.15) K = 2051.15 K
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (3+1)-(2) = 2
Thus, Kp is:
Answer:
0.00 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ 30.07 32.00
2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O
Mass/g: 1.50 11.
2. Calculate the moles of each reactant
3. Calculate the moles of CO₂ we can obtain from each reactant
From ethane:
The molar ratio is 4 mol CO₂:2 mol C₂H₆
From oxygen:
The molar ratio is 4 mol CO₂:7 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is ethane, because it gives the smaller amount of CO₂.
The excess reactant is oxygen.
5. Mass of ethane left over.
Ethane is the limiting reactant. It will be completely used up.
The mass of ethane left over will be 0.00 g.
Answer:
Pressure gas A
using boyles law
=
V2
V2 = 717ml + 179 ml
= 896ml
∴ = 2.50 × 717ml/896ml
= 2.0 bar
Pressure B
P2 = 4.30 bar× 179ml/896ml
= 0.859bar
ptotal = +
= 2.0 bar + 0.859 bar
= 2.859 bar
Explanation: Using Daltons law of partial pressure,the pressure is independently of each other when the gas is exerted.where we can use daltons law to find the pressure of each gas separately when it expands into the total volume in two containers.