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Maru [420]
2 years ago
7

For the following reaction, find the value of Q and predict the direction of change, given that a 1L flask initially contains 2

moles S8, 2 moles SF6, and 2 moles F2.
1/8 S8 (s) + 3 F2 (g) ⇄ SF6 (g) Kc = 0.425

(A) Q = K, system is at equilibrium
(B) Q < K, reaction will make more reactants
(C) Q < K, reaction will make more products
(D) Q > K, reaction will make more reactants
(E) Q > K, reaction will make more products
Chemistry
1 answer:
Tresset [83]2 years ago
5 0

Answer:

C) Q < K, reaction will make more products

Explanation:

  • 1/8 S8(s)  + 3 F2(g)  ↔  SF6(g)

∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

∴ [ SF6 ] = 2 mol/L

∴ [ F2 ] = 2 mol/L

⇒ Q = ( 2 ) / ( 2³)

⇒ Q = 0.25

⇒ Q < K, reaction will make more products

 

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0.53313648 feet = 16.25cm

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3 years ago
What are the names of the stable forms of oxygen?
zloy xaker [14]

Answer:

18 O, 17 O, and 16 O

Explanation:

three naturally stable isotopes

5 0
2 years ago
A chemical reaction is shown below:
BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

Mass of glucose = 20 g

Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂     →   6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

               C₆H₁₂O₆           :            H₂O

                   1                   :              6

                 0.11                :           6/1×0.11 = 0.66

                   O₂               :            H₂O

                   6                   :              6

                 0.47                :           0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol  ×18 g/mol

Mass = 8.46 g

8 0
3 years ago
Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,
storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

Cu(s) + S\rightarrow CuS

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

moles=\frac{54}{63.55}

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

moles=\frac{42}{32.06}

Moles of S = 1.31 mol

Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n

<u><em>First find the limiting reagent :</em></u>

Cu + S\rightarrow CuS

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require  = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required  = 0.8497 mol

S is present in excess and <u>Cu is limiting reagent</u>

<u>All Cu is consumed in the reaction</u>

Amount Cu will decide the amount of CuS formed

Cu + S\rightarrow CuS

1 mole of Cu  gives = 1 mole of Copper sulfide

0.8497 mol of Cu =  1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

Moles = \frac{given\ mass}{Molar\ mass}

0.8497 = \frac{given\ mass}{95.611}

Mass of CuS = 0.8497 x 95.611

= 81.24 g

3 0
3 years ago
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BabaBlast [244]

Explanation:

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7 0
3 years ago
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