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3 years ago

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An astronomy research team has been studying the atmosphere of a moon orbiting a newly discovered exoplanet. The team has determ

ined that the moon has an average temperature of 95K on the surface, with an average pressure of 1.6atm. Remote analysis of this moon's atmosphere has revealed it has a molar mass of 28.6 g/mol. Calculate the density (g/L) of 1 mole of the moon's atmosphere under the given conditions.

1 answer:

sineoko [7]3 years ago

8 0

Answer:

5.81 g/L

Explanation:

Let's apply the Ideal Gases Law to determine this:

P . V = n . R . T

Pressure = 1.6 atm

Volume = ?

Mol = 1 mol

Temperature = 96 K

In order to find the density, we should know the volume of the atmosphere which is a mixture of gases so, we consider all the atmosphere as a unique ideal gas.

1. 6 atm . V = 1 mol . 0.082 L.atm/mol.K . 96K

V = (1 mol . 0.082 L.atm/mol.K . 96K) / 1.6 atm

V = 4.92 L → As this is the volume for the whole atmosphere and the mass of 1 mol is 28.6 g, density should be:

28.6 g / 4.92L = 5.81 g/L

Density → mass / volume

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A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

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

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

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4 years ago

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PSYCHO15rus [73]

Explanation:

(a)

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength
$

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Given, Wavelength = 505 nm

Also, 1 m = $10^{-9}$ nm

So,

Wavelength = $505\times 10^{-9}$ m

Thus, Frequency is:

$Frequency=\frac{c}{Wavelength}$

$Frequency=\frac{3\times 10^8}{505\times 10^{-9}}\ Hz$

$Frequency=5.94\times 10^{14}\ Hz$

(b)

Also, $E=\frac {h\times c}{\lambda}$

Where,

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So,

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{505\times 10^{-9}}\ J$

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In order of the increasing frequency and the photon energy and the decreasing wavelength the spectrum are:

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Alisiya [41]

Answer:Hence, the bond length in HCl is 125 pm.

Explanation:

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Also given that bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter-nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Beyond this if atoms come closer to each other then their will be repulsion between them.

So, the bond length between the Hydrogen and Chlorine atom in HCl molecule is :

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3 years ago

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pochemuha

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dybincka [34]

Answer:

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3 years ago