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Naily [24]
1 year ago
15

The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C. 2NO (g) + O2 (g) 2NO2 (g) What is the value of Keq at this temperat

ure for the following reaction? NO2 (g) NO (g) + 1/2 O2 (g)
Chemistry
1 answer:
kodGreya [7K]1 year ago
5 0

<u>Answer:</u> The equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

<u>Explanation:</u>

The given chemical equation follows:

2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)

The value of equilibrium constant for the above equation is K_{eq}=5.4\times 10^{13}

Calculating the equilibrium constant for the given equation:

NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)

The value of equilibrium constant for the above equation will be:

K'_{eq}=\frac{1}{\sqrt{K_{eq}}}\\\\K'_{eq}=\frac{1}{\sqrt{5.4\times 10^{13}}}\\\\K'_{eq}=1.36\times 10^{-7}

Hence, the equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

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Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre
vaieri [72.5K]

Answer:

The pressure 10,000 m below the surface of the sea is 137.14 MPa.

The density 10,000 m below the surface of the sea is 2039 kg/m3

Explanation:

P0 and ρ0 are the pressure and density at the sea level (atmosferic condition). As the depth of the sea increases, both the pressure and the density increase.

We can relate presure and density as:

\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\

Rearranging

\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta}  =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\

With this equation, we can calculate P at 10,000 m below the surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density at 10,000 m below the surface of the sea is

\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3

4 0
2 years ago
2. Newton began his academic career in 1667. For how long was he a working scientist? Was he a very productive scientist? Justif
Alla [95]
The title of Scientist was formally bestowed upon Sir Isaac Newton when he was
awarded the Merit Badge in Science at the age of 15, and he remained a Scientist
until he died, at the age of 84, on March 20, 1727, for a total duration of 69 years.

Yes, He was a productive scientist.He is one of the most important contributors to our understanding of how the universe works.
5 0
1 year ago
You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point
Hoochie [10]
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
 x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

7 0
1 year ago
Read 2 more answers
When a 3.22 g sample of an unknown hydrate of sodium sulfate, na2so4 ⋅ h2o(s), is heated, h2o (molar mass 18 g) is driven off. T
Mariana [72]

The value of X is 10 hence the formula of unknown hydrate   sodium sulfate is  NaSO4.10 H20

calculation

step 1:find the moles of NaSO4 and the moles of H2O

moles= mass/molar mass

moles of Na2SO4=1.42÷142=0.01 moles

moles of H20=  mass of H2O/molar mass of H2O

mass of H2O= 3.22-1.42=1.8g

mole of H2O is therefore 1.8÷18=0.1 moles

step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)

that is Na2So4= 0.01/0.01 =1

            H2O= 0.1/0.01=10

6 0
2 years ago
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3
ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×\frac{1molZn}{65,38g}×\frac{1molH_{2}(g)}{1molZn} = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×\frac{101,325}{1atmL} = <em>19,26 kJ</em>

I hope it helps!

5 0
2 years ago
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