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dexar [7]
2 years ago
5

Each of the chemically active Period 2 elements forms stable compounds in which it has bonds to fluorine.(d) Draw Lewis structur

es for these compounds.
Chemistry
1 answer:
Pavel [41]2 years ago
5 0

Lithium is the first element of period 2 which reacts with fluorine to form LiF ( lithium fluoride ) . it is an inorganic compound . it is also a colorless solid . it is less soluble in water . it is chemically stable because of its comparable molecular mass .

Beryllium is the second element of period 2 which reacts with fluorine to give beryllium difluoride (BeF2) . it is inorganic compound . it is highly soluble in water. it is also a stable compound . it have low melting point .

Boron is the third element of period 2 which reacts with fluorine to form

BF3 (Boron trifluoride ) . it is a inorganic compound . it is colorless and toxic gas forms  . it is stable in dry atmosphere but its octet is not satisfied .

Carbon is the 4th element of the period 2 which reacts with fluorine to form carbon tetrafluoride (CF4) . it is not soluble in water . it is a greenhouse gas . it dissolves in oil. it is very stable compound .it forms covalent bond .

Nitrogen is the 5th element of period 2 which reacts with fluorine to form nitrogen trifluoride (NF3) . it is also a inorganic compound . it  is colorless and non-flammable .  it is a stable gas at room temperature .

Oxygen is the 6th element of period 2 which reacts with fluorine to form oxygen difluoride (OF2) . it is colorless poisonous gas . it is partially stable or relatively stable .

Neon is a noble gas and also a stable element . it is odorless and colorless . so it is nonreactive . so it doesn't form bond with fluorine .

<h3>Learn more about Fluorine here :</h3>

brainly.com/question/3494441

#SPJ4

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3 years ago
The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g) Calculate the equili
dimaraw [331]

Answer:

pCH4 =  0.9184 atm

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Explanation:

Step 1: Data given

The equilibrium constant, Kp= 9.52 * 10^-2

Temperature = 350 K

Each have an initial pressure of 1.06 atm

Step 2: The balanced equation

CH4(g) + CCl4(g) ⇆ 2CH2Cl2(g)

Step 3: The pressure at the equilibrium

pCH4 = 1.06 - X atm

pCCl4 = 1.06 - X atm

pCH2Cl2 = 2X

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X = 0.1416

Step 5: Calculate the partial pressure

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pCCl4 = 1.06 - 0.1416 =  0.9184 atm

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Kp = (0.2832²) / (0.9184*0.9184)

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pCH2Cl2 = 0.2832 atm

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ON

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MA = \frac{5}{0.75}

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