The change in the internal energy of the gas is 1.5×10∧3 J.
The internal energy of an ideal gas is directly proportional to the temperature of the gas:
ΔE = 3/2 × n × R × ΔT
ΔT = 320 K - 260 K
ΔT = 60 K; change of the temperature
n = 2.0 mol: amount of a monatomic ideal gas
R = 8.1 J/mol×K;the ideal gas constant
ΔE = 3/2 × 2 mol × 8.1 J/mol×K × 60 K
ΔE = 1500 J
ΔE = 1.5×10∧3 J; the internal energy of the gas
Isobaric process is a type of process in which the pressure of the system stays constant.
More about an isobaric process: brainly.com/question/28106078
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Low life?- i’m pretty sure
Answer:A. Increases
Explanation:
Heating or an increase in temperature increases the kinetic energy of particles thereby increasing their motion and how they relate and react with one another.
Increase in the temperature of the solvent is directly proportional to the rate of dissolution. The rate of dissolution increases due to the increase in kinetic energy. This makes the solvent particles interact faster with the solute particles thereby increasing the dissolution rate.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
Fair Labor Standards Act
Explanation:
It is correct. I just took the test on Clever
