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3 years ago

If the vapor pressure of ethanol at 34.7degree C is 100

Chemistry

1 answer:

saw5 [17]3 years ago

6 0

Answer:

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

ln (P2/P1) = $\frac{ΔvapH}{R}([tex]\frac{1}{T1}$ - $\frac{1}{T2}$ )

where

P1 and P2 are the vapour pressures at temperatures T1 and T2

Δ vapH = the enthalpy of vaporization of the ETHANOL

R = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

; T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

0.0004368=(0.00325 - T⁻²)

T⁻²=0.002813

T² = 355.47K

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Elza [17]

1) 2P

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Explanation:

To solve the first part of the problem, we can use Boyle's Law, which states that:

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Mathematically:

$pV=const.$

where

p is the pressure of the gas

V is its volume

For the Xe gas in this problem we can write

$p_1 V_1 = p_2 V_2$

where:

$p_1 = P$ is the initial pressure

$V_1=V$ is the initial volume

$V_2=\frac{V}{2}$ is the final volume

Solving for p2, we find the final pressure:

$p_2=\frac{P V_1}{V_2}=\frac{pV}{V/2}=2P$

So, the pressure has doubled.

The average speed of the atoms/molecules in the gas is given by the formula

$v_{rms}=\sqrt{\frac{3RT}{M}}$

where

R is the gas constant

T is the absolute temperature (in Kelvin) of the gas

M is the molar mass of the gas

$v_{rms}$ is known as rms speed of the particles in the gas

From the formula, we see that the speed of the atoms in the gas depends only on the temperature of the gas.

In the Xe(g) gas in this problem, the temperature is kept constant; therefore, since nothing changes in the formula, this means that the average speed also does not change.

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3 years ago

The average temperature of healthy person is 98.6°F. What is it in Celsius scale?

meriva

Answer:

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Explanation:

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2 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

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$E_n=-13.6\times \frac{Z^2}{n^2}ev$

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Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

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3 years ago

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miss Akunina [59]

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