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2 years ago

A 2 L container contains 5 moles of helium gas at 0.5 atm. What is the temperature of the gas?

Chemistry

1 answer:

lara [203]2 years ago

7 0

Answer:

2.44 K IS THE TEMPERATURE OF THE GAS

Explanation:

PV = nRT

P = 0.5 atm

V = 2 L

n = 5 moles

R = 0.082 L atm mol^-1 K^-1

T = ?

Substituting for T in the equation, we obtain:

T = P V / nR

T = 0.5 * 2 / 5 * 0.082

T = 1 / 0.41

T = 2.44 K

The temperature of the gas is 2.44 K

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Consider the following reaction at equilibrium. What effect will decreasing the temperature have on the system? DH=+890kJCO2(g)+

AlladinOne [14]

Answer:

Option B. The reaction will shift to the left in the direction of the reactants.

Explanation:

The equation for the reaction is given below:

CO₂ + 2H₂O <=> CH₄ + O₂

Enthalpy change (ΔH) = +890 KJ

The reaction illustrated by the equation is endothermic reaction since the enthalpy change (ΔH) is positive.

Increasing the temperature of an endothermic reaction will shift the equilibrium position to the right and decrease the temperature will shift the equilibrium position to the left.

Therefore, decreasing the temperature of the system illustrated by the equation above, will shift the reaction to the left in the direction of the reactants.

Thus, option B gives the right answer to the question.

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Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm

ivolga24 [154]

Answer:

$O_{2}$ and $NO_{2}$

Explanation:

For a given system at constant temperature, the number of moles of gas present in the system is proportional to the product of the system pressure and volume. Therefore, we have:

NO: 6 L * 0.7 atm = 4.2 L*atm

O: 1.5 L* 2.5 atm = 3.75 L*atm

For the given system based on a balanced chemical equation:

2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e. $P_{NO}$ = 0

For the product: 2.70 L*atm NO produced 2.70 L*atm $NO_{2}$

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of $NO_{2}$ = (2.70 L*atm $NO_{2}$ ) / (7.5 L) = 0.36 atm $NO_{2}$

Similarly for oxygen gas:

3.75 L*atm - 1.35 L*atm = 2.40 L*atm oxygen gas remaining

Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm

Thus, the gases present at the end of the experiment are $O_{2}$ and $NO_{2}$

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2 years ago