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ArbitrLikvidat [17]
3 years ago
9

For the reaction below at a particular temperature, the concentrations at equilibrium were observed to be: [F2] = 2.1× 10–2 mol/

L and [F] = 2.0 × 10–4mol/L. Calculate the value of the equilibrium constant from these data. (The units are deleted.)
Chemistry
1 answer:
grin007 [14]3 years ago
5 0

Answer:

Kc = 1.9x10^-6

Explanation:

You are missing the reaction, however, let me write the reaction for you:

F2(g) <----------> 2F(g)

Now, remember that for an equilibrium reaction, that means that a reaction is not complete at all, we have a value of equilibrium constant, usually called Kc.

This value of Kc is the result of concentrations of products and reactants:

Kc = [Products] / [Reactants]

However, you should note that only substances and compounds in gaseous phase and aquous contribute to the equilibrium. This is because solid and liquid concentrations values are constants and near to 1.

Now for a general reaction like this one:

aA(g) + bB(g) <--------> cC(g) + dD(g)

The expression of Kc would be:

Kc = [C]^c * [D]^d / [A]^a * [B]^b

The letters a, b, c, d are the coefficients that balance the equation and they should be considered in the equilibrium, that is why they are elevated.

Now, with this clear, let's apply these concepts to the reaction here:

F2(g) <----------> 2F(g)

The expression of Kc will be:

Kc = [F]² / [F2]

Replacing the given values in the above expression, we can solve for Kc:

Kc = (2x10^-4)² / (2.1x10^-2)

Kc = 1.9x10^-6

This is the value of Kc and have no units.

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Answer:

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7 0
3 years ago
How many moles of Fluorine (F2) are<br><br> needed to completely react 8.0<br><br> moles of NF3?
Anna [14]

Answer:

12 moles of F₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e

Xmol of F₂ = (3 × 8)/2

Xmol of F₂ = 12 moles

Thus, 12 moles of F₂ is needed for the reaction.

8 0
3 years ago
What is the percent composition of N H S O in (NH4)2SO4
Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

- 1 mole of S (32g/mol) = 32g

- 4 moles of O (16g/mol) = 64g

3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\  \\  \end{gathered}

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

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