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Nata [24]
4 years ago
7

How many millibars are equal to 27:23 inches of mercury. How did you get you answer

Chemistry
1 answer:
Nady [450]4 years ago
4 0

the correct answer is.. ill tell you when i search it up gimme 2 seconds


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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

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

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Blizzard [7]

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

8 0
3 years ago
Which one of the following represents a decrease in entropy?
Leni [432]

Answer:

D

Explanation:

There are principally three states of matter. These are the solid, liquid and gaseous states. The gaseous state has the highest degree of disorderliness as gas particles can move randomly while the solid state has the highest level of compactness.

Hence, we need to be adequately fed with information as regards the phase change to know if entropy has decreased or increased.

A. is wrong

Evaporation is a change of state to the gaseous state meaning there is an increased entropy.

B. is wrong

Sublimation is a change of state which means a solid substance like iodine or naphthalene changes its state directly to the gaseous state. There is an increased entropy here too.

C is wrong

Melting of ice means going from ice block to liquid water. This is synonymous to going from the solid state to the liquid state which is an increased entropy

D is correct

Condensation involves going from the gaseous state to the liquid state. This means going from a less ordered state to a more ordered state. This is accompanied by an entropy decrease.

E is wrong

While there are some processes that increase entropy, we also have some process that decrease entropy.

7 0
3 years ago
He isotope 62(over28ni has the largest binding energy per nucleon of any isotope. calculate this value from the atomic mass of n
irina [24]
The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 =  0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.
4 0
3 years ago
The half-life of 131I is 80 days. How much of a 500.0 mg sample remains after 3 half lives?
I am Lyosha [343]

Answer:

62.5 mg

Explanation:

Just multiply the original amount by  1/2   three times:

500 mg  x  1/2  x  1/2  x  1/2 = 62.5 mg

6 0
2 years ago
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