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Nookie1986 [14]
3 years ago
7

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction2COF2(g)⇌CO2(g)+CF4(g), Kc=5.70If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Chemistry
1 answer:
Anna11 [10]3 years ago
7 0

Answer:

[COF₂] = 0.346M

Explanation:

For the reaction:

2COF₂(g) ⇌ CO₂(g) + CF₄(g)

Kc = 5.70 is defined as:

Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 <em>(1)</em>

Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:

[COF₂] : 2.00M - 2x

[CO₂] : x

[CF₄] : x

Replacing in (1):

5.70 =  [X] [X] / [2-2x]²

22.8 - 45.6x + 22.8x² = x²

0 = -21.8x² + 45.6x - 22.8

Solving for x:

X = 1.265 <em>-False answer, will produce negative concentrations-</em>

<em>X = 0.827.</em>

Replaing, molar concentration of COF₂ is:

[COF₂] : 2.00M - 2×0.827 = <em>0.346M</em>

I hope it helps!

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What ion (cation or anion and charge) do the halogen group (17) form to fulfill the octet
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Answer:

option B= anion (-1)

Explanation:

The halogens are present in the P- block of periodic table. There are thirty five elements present in p-block. The halogens are present in group seventeen. Their are seven electrons are present in valance sheet. All halogens require one electron to complete the octet and form anion carry -1 charge.

Halogens include following elements:

Fluorine, chlorine, bromine, iodine and astatine.

The halogens mostly form single bond with the other elements by gaining one electron and -1 charge. These are most reactive non-metals.

Properties of some halogen elements:

Fluorine:

1. It is yellow in color.

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3. It is highly corrosive.

4. Fluorine has pungent smell.

5. It's reactions with all other elements are vigorous except neon, oxygen, krypton and helium.  

Chlorine:

1. It is greenish-yellow irritating gas.

2. Its melting point is 172.2 K.

3. Its boiling point is 238.6 K.

4. It is disinfectant and can kill the bacteria.

5. It is also used in manufacturing of paper, paints and textile industries.

Iodine:

1. It is present in solid form.

2. It is crystalline in nature.

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3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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