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Anastaziya [24]
3 years ago
6

Consider the following chemical equilibrium:

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

Kc = [CH₄] / [H₂]²

Kp = [CH₄] / [H₂]² * (0.082*T)^-1

Explanation:

Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.

<em>Pure solids and liquids are not taken into account in an equilibrium</em>

Thus, for the reaction:

C(s)+ 2H₂(g) ⇌ CH₄(g)

Equilibrium constant is:

<h3>Kc = [CH₄] / [H₂]²</h3>

Now, using the formula:

Kp = Kc* (RT)^Δn

<em>Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)</em>

Replacing:

<h3>Kp = [CH₄] / [H₂]² * (0.082*T)^-1</h3>

<em />

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rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
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2 years ago
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Answer:

I think it is b.

Explanation:

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Lab report on flock flock y flock z insects and fruits
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2 years ago
Calculate the mass of calcium chloride that contains 3.20 x 1024 atoms of chlorine.
frez [133]

Answer:

294.87 gm  CaCl_2

Explanation:

The computation of the  mass of calcium chloride is shown below:

But before that following calculations need to be done

Number of moles of chlorine atom is

= 3.20 × 10^24 ÷ 6.022 × 10^23

= 5.314 moles

As we know that

1 mole CaCl_2 have the 2 moles of chlorine atoms

Now 5.341 mole chloride atoms would be

= 1 ÷ 2 × 5.314

= 2.657 moles

Now

Mass of CaCl_2 = Number of moles × molar mass of  CaCl_2

= 2.657 moles × 110.98 g/mol

= 294.87 gm  CaCl_2

7 0
3 years ago
18. How many moles of atoms are there in each of
Lelechka [254]

a) 1 mole of Ne

b) i/2 mole of Mg

c) 1570 moles of Pb.

d) 2.18125*10^-13 moles of oxygen.

                     

Explanation:

The number of moles calculated by Avogadro's number in 6.23*10^23 of Neon.

6.23*10^23= 1/ 6.23*10^23

                   = 1 mole

The number of moles calculated by Avogadro's number in 3.01*10^23 of  Mg

3.2*10^23=1/6.23*10^23

                = 1/2 moles of Pb.

Number of moles in 3.25*10^5 gm of lead.

atomic weight of Pb=

n=weight/atomic weight

  = 3.25*10^5/ 207

  = 1570 moles of Pb.

Number of moles 4.50 x 10-12 g O

number of moles= 4.50*10^-12/16

                            =  2.18125*10^-13 moles of oxygen.

3 0
3 years ago
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