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Zielflug [23.3K]
2 years ago
13

Given equilibrium partial pressures of PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm calculate the equilibrium constant

. Which direction does the reaction need to go to reattain equilibrium if the current reaction partial pressures are PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm?
Chemistry
2 answers:
maxonik [38]2 years ago
5 0
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.

In present case, following reaction is involved:

                        2NO2    ↔      2NO + O2

Here, K = \frac{[PNO]^2[O2]}{[PNO2]^2}

Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence,  K = \frac{[0.0022]^2[0.0011]}{[0.247]^2}
                 = 8.727 X 10^-8

Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................

Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.

Therefore, Reaction quotient = </span>\frac{[PNO]^2[O2]}{[PNO2]^2}
                                              = \frac{[0.021]^2[0.037]}{[0.192]^2}
                                              = 4.426 X 10^-4.

Here, Reaction quotient > Equilibrium constant.

Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>
Dominik [7]2 years ago
3 0

Answer:toward the reactants

Explanation:

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In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

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6 0
2 years ago
Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta
Arlecino [84]

Answer:

A. \lambda_0=2.196\times 10^{-7}\ m

Explanation:

The work function of the Platinum = 9.05\times 10^{-19}\ J

For maximum wavelength, the light must have energy equal to the work function. So,

\psi _0=\frac {h\times c}{\lambda_0}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda_0 is the wavelength of the light being bombarded

\psi _0=Work\ function

Thus,

9.05\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda_0}

\frac{9.05}{10^{19}}=\frac{19.878}{10^{26}\lambda_0}

9.05\times \:10^{26}\lambda_0=1.9878\times 10^{20}

\lambda_0=2.196\times 10^{-7}\ m

8 0
2 years ago
A lead cylinder has a mass of 540 grams and a density of 2.70 g/ml. What is its volume
ch4aika [34]
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Volume = 540g / 2.70 g/ml

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3 0
2 years ago
Read 2 more answers
2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to
Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode (m_{ch}) is given as :

\frac{m_{ch}}{M} ZF = It

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; (m_{ch}) = (Ad) \rho ; replacing that into above equation; we have:

\frac{(Ad) \rho}{M} ZF = It  ---- equation (1)

where;

A = area

d = thickness

\rho = density

From the above equation (1); The time required for coating can be calculated as;

[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons }  ] = (20 A) t

t = \frac{2100}{20}

= 105 s

8 0
2 years ago
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