What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)

Answers

Answer 1

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Best regards!


Related Questions

When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation

Answers

During sublimation

This has been posted on here before so you could’ve searched it lol.

Best of luck :))

Answer:

during sublimation

Explanation:

just took the test

What can the chemical formula tell us about a compound?

Answers

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

Hope this helps! please mark me brainliest!

God bless :)

SOMEONE PLEASE HELPPP​

Answers

A is the correct awnser Beacuse it ether right kne

A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.

a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g

Answers

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

What is empirical formula?

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

To learn more about empirical formula, refer to the link below:

https://brainly.com/question/11588623

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