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liberstina [14]
3 years ago
7

What is deltaG at 298 K for the reaction C + O2 = CO

Chemistry
1 answer:
blondinia [14]3 years ago
8 0

The answer for the following problem is mentioned below.

Therefore the Gibbs free energy (ΔG) for the reaction is <u><em>-394.36 J</em></u>

Explanation:

Given:

ΔH = -393.5 kJ/ mol

ΔS =0.0029 kJ/mol K

T = 298 K

To solve:

Gibbs free energy (ΔG)

We know;

<u>ΔG = ΔH - TΔS </u>

ΔG  = -393.5 - (298 × 0.0029)

ΔG = -393.5 - 0.8642

<u><em>ΔG = -394.36 J</em></u>

Therefore the Gibbs free energy (ΔG) for the reaction is <u><em>-394.36 J</em></u>

<u><em></em></u>

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Answer:

6 electrons

Explanation:

The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s.

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3 years ago
What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?
White raven [17]

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

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5 0
3 years ago
What does velocity describe
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7 0
3 years ago
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The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

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3 years ago
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