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Deffense [45]
3 years ago
8

Consider the following reversible reaction

Chemistry
1 answer:
kogti [31]3 years ago
7 0

Answer:

The right choice is  K(eq) = ([CO₂]) / [O₂]

Explanation:

The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium

For a system undergoing a reversible reaction described by the general chemical equation

  • a A + b B → x X + n N

An equilibrium constant can be expressed as following

K(eq) = ([X]ˣ*[N]ⁿ) / ([A]ᵃ*[B]ᵇ)

where [] is the equilibrium concentration of each component.

and the concentration of solid is conventionally considered as 1

<u>in your example: </u>

The reversible reaction

  • C ₍s₎ + O₂ ₍g₎  ↔ CO₂ ₍g₎

taking in consideration that C is a solid and the concentration of solid is conventionally considered as 1

So,  the equilibrium constant for this reversible reaction can be expressed as following

  • K(eq) = ([CO₂]) / [O₂]

∴ The right choice is  K(eq) = ([CO₂]) / [O₂]

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The correct answer would be... Scientific
8 0
3 years ago
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
3 years ago
How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq
Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

8 0
3 years ago
Type the correct answer in the box. The pressure inside a tire is measured as 28.0 . What is its pressure in ? 1 pound = 4.45 ne
posledela

The pressure is 19.3 N/cm².

<em>p</em> = (28.0 lb/1 in²) × (4.45 N/1 lb) × (1 in²/6.45 cm²) = 19.3 N/cm²

8 0
4 years ago
Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
 (12/44) × 8.46 = 2.3073 g
 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
4 years ago
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