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3 years ago

The equation of state corresponding to 8 gm of Oxygen (O2) is : PV=RT,RT/4,RT/3 or RT/2?

Chemistry

2 answers:

nexus9112 [7]3 years ago

8 0

In PV = nRT, n is the number of moles. When 8 grams of oxygen are present,
n = 8/32 = 1/4
So PV = RT/4

Stels [109]3 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

According to ideal gas equation, the product of pressure and volume is equal to the product of universal gas constant and absolute temperature.

Mathematically, PV = nRT ........... (1)

where P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

Also, number of moles = $\frac{mass of atom}{molar mass of the atom}$

Therefore, number of moles of oxygen atom will be as follows.

Number of moles = $\frac{mass of atom}{molar mass of the atom}$

= $\frac{8}{32}$

= $\frac{1}{4}$

Thus, putting the value of n into equation (1) as follows.

PV = nRT

or PV = $\frac{1}{4}RT$

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Which of the following statements is incorrect? Group of answer choices The product of wavelength and frequency of electromagnet

morpeh [17]

Answer: Option (b) is the correct.

Explanation:

It is known that energy is directly proportional to frequency.

Mathematically, $E \propto \nu$

or, E = $h \times \nu$ ............ (1)

where, E = energy

h = Plank's constant

$\nu$ = frequency

This relation shows that higher is the energy then more will be the frequency of electrmagnetic radiation.

Also, $\nu = \frac{c}{\lambda}$ ......... (2)

where, c = speed of light

$\lambda$ = wavelength

Therefore, substituting value of $\nu$ from equation (2) into equation (1) as follows.

E = $h \times \nu$

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

This relation shows that higher is the energy then lower will be the wavelength.

Thus, we can conclude that as the energy of a photon increases, its frequency decreases, is incorrect statement.

3 0

3 years ago

Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.

Vsevolod [243]

Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06

5 0

3 years ago

Read 2 more answers

In a classical long run supply model the economy is always doing what

ch4aika [34]

Answer:

Economy is always at the full employment level of output

Explanation:

The economy in a classical long-run supply model will always have the same economic output

8 0

3 years ago

Please tell if my answer is right or wrong.