Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomi
c gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are diff erent. Determine the ratio KEKrypton KEArgon of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom.
1 answer:
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2.168 L of air will leave the container as it warms
<h3>Further explanation</h3>Given
V₁=2.05 L
T₁ = 5 + 273 = 278 K
T₂ = 21 + 273 = 294 K
Required
Volume of air
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature
Input the value :
V₂=(V₁.T₂)/T₁
V₂=(2.05 x 294)/278
V₂=2.168 L
Explanation:
d.
Answer : The enthalpy of the reaction is, -2552 kJ/mole
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given enthalpy of reaction is,
The intermediate balanced chemical reactions are:
(1)
(2)
(3)
(4)
Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :
(when we are reversing the reaction then the sign of the enthalpy change will be change.)
The expression for enthalpy of the reaction will be,
Therefore, the enthalpy of the reaction is, -2552 kJ/mole
Answer:
3.69 g
Explanation:
Given that:
The mass m = 325 g
The change in temperature ΔT = ( 1540 - 165)° C
= 1375 ° C
Heat capacity = 0.490 J/g°C
The amount of heat required:
q = mcΔT
q = 325 × 0.490 × 1375
q = 218968.75 J
q = 218.97 kJ
The equation for the reaction is expressed as:
Then,
1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.
Thus, for 218.97 kJ, the amount of ethyne gas required will be:
= 3.69 g