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avanturin [10]
3 years ago
9

What is the internal energy u of one mole of air on a very hot summer day (35∘c)? express your answer numerically in joules to t

wo significant figures?
Chemistry
1 answer:
umka21 [38]3 years ago
6 0
- For solving this problem, we have to take in account the degree of freedom of air molecules.
- As, molecules of air have five degrees of freedom (three translational and two rotational)
- For each molecule, the kinetic energy of each degree of freedom = 1/2 KT
- So, K.E of 1 molecule for 5 degrees of freedom = 5/2KT
So, for molecules of air K.E = 5/2 KT = 5/2 x 1 x 1.38 x 10⁻²³ x 308 = 1062.6 x 10⁻²³ J
1 mole of air contains 6.022 x 10²³ molecules 

K.E. of 1 mole = 1062.6 x 10⁻²³ x 6.022 x 10²³ = 6400 J
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HELP!! Suppose that the pressure of 1.00 L of gas is 380 mm Hg when the temperature is 200. K. At what
Korolek [52]
Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.



sendo:

τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)


temos: τ = Q1 - Q2

O rendimento (η) é a razão do trabalho realizado pela energia fornecida:

η= τ/Q1

Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.

solução:

τ=3500 J
Q1=5000J

η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%

Energia dissipada será:



τ = Q1 - Q2
Q2 = Q1- τ

Q2=5000-3500
Q2= 1500 J

Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?



obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem
4 0
3 years ago
When making calculations, you should rely on the precision of your measured data.
ELEN [110]
True 

Im happy i could help you today 

have a great rest of ur week :)
3 0
3 years ago
Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)
Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}  

1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%                      

   

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

With [CH₃COOH] = 0.01 M

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}  

1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!    

4 0
3 years ago
Which transition by an electron will release the greatest amount of energy? answer soon please ​
alexira [117]

Answer:

When electron jumps from high energy level to lower energy level.

Explanation:

The elctronic transition from one energy level to another energy level within the atom, always involve energy transitions.

The energy released or absorbed by electronic transition is always discrete and is called as " Photon". It means when electron jumps from when energy level to another energy level the energy released or absorbed is treated as photon emitted or absorbed.

When an electron jumps from higher energy level to a lower energy level, a photon of specific wavelength and specific energy is emitted in other words we can say that energy is released or emitted.

The energy of photon emitted or absorbed is easily calculated using Rydberg Formula which is simply the energy difference between the two energy levels and is given as under;

Ephoton = Eo ( 1 / n1^2 - 1 / n2 ^ 2)

In the above formula n1 is the initial energy level of electron and n2 is the final energy level of electron.

Eo = 13.6 eV ( Here "o" in Eo is in subscripts)

In n1 and n2 1 and 2 are in the subscripts.

^ represents that the disgits after them are exponents.

So by just putting the values of energy levels n1 and n2 we can easily calculate the value of energy of photon ( energy due to electronic transition) and compare the results that which transition will give high energy photon and which will give low energy photon.

3 0
4 years ago
What is the primary source of gas for thick terrestrial planet atmospheres?
marishachu [46]
Outgassing from active volcanoes
7 0
4 years ago
Read 2 more answers
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