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pishuonlain [190]
3 years ago
6

A gas at constant volume has a pressure of 3.20 atm at 300. K. What will be the pressure of the gas at 290. K?

Chemistry
1 answer:
melomori [17]3 years ago
3 0
  • From the general law of gases: PV = nRT,

where P is the pressure (atm),

V is the volume (L),

n is the number of moles,

R is the general gas constant (8.314 L.atm/mol.K),

T is the temperature in Kelvin

  • at constant volume of the gas: P1T2 = P2T1

P1 = 3.20 atm, T1 = 300 K, T2 = 290 K, P2 = ??

(3.20 atm)(290 K) = P2(300 K)

P2 = (3.20 atm)(290 K)/ (300 K) = 3.093 atm

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How can a substance stay at a certain temperature even though it is being heated?​
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I'm not 100% sure of the answer. Explanations are highly appreciated.
PolarNik [594]

Answer:

The correct answer is

B.

\Delta H=2(1072)+498 -4(799)

Explanation:

Enthalpy of reaction :

It is the amount of energy released/absorbed when one mole of the substance is formed from the reactant at a constant pressure.

The enthalpy of a reaction can be calculated using :

\Delta H=\Delta H_{reactants}-\Delta H_{products}

2CO+O_{2}\rightarrow 2CO_{2}

\Delta H_{reactants}=2(C\equiv O)+O=O

\Delta H_{reactants}=2(1072)+498

H_{products}=2(2\times C=O)

H_{products}=4\times 799

\Delta H=\Delta H_{reactants}-\Delta H_{products}

\Delta H=2(1072)+498 -4(799)

units = \frac{kJ}{mole}

Please note that :

The carbon monoxide , CO should be taken as C triple bond O. Not C=O .

So , the bond energy =1072 is used

\Delta H=2(1072)+498 -4(799)

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