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kolezko [41]
3 years ago
7

Chlorine gas (cl2) forms when two chlorine atoms share an electron. what type of bonding is present in chlorine gas?

Chemistry
2 answers:
Masja [62]3 years ago
7 0
Is there some choises
zhenek [66]3 years ago
5 0

Explanation:

When a chemical bond is formed by sharing of electrons then it is known as a covalent bond.

A covalent bond is always formed between two non-metals.

For example, atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7.

As a result, it needs one more electron to complete its octet. So, when it chemically combines with another chlorine atom then there will occur sharing of electrons.

Hence, Cl_{2} is formed which is a covalent compound.

On the other hand, when a chemical bond is formed by transfer of electrons from one atom to another then it is known as an ionic bond. An ionic bond is always formed between a metal and a non-metal.

For example, LiCl is an ionic compound.

Thus, we can conclude that in a chlorine gas covalent bonding is present when two chlorine atoms share an electron.

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The radius of a lithium atom is 130 picometers, and the radius of a fluorine atom is 60 picometers. The radius of a lithium ion,
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During a hot summer, air temperatures along a coastline are higher above the land than they are above the water, even though bot
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The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.
Kitty [74]

Answer:

a) The theoretical yield is 408.45g of BaSO_{4}

b) Percent yield = \frac{realyield}{408.45g}*100

Explanation:

1. First determine the numer of moles of BaCl_{2} and Na_{2}SO_{4}.

Molarity is expressed as:

M=\frac{molessolute}{Lsolution}

- For the BaCl_{2}

M=\frac{1.75molesBaCl_{2}}{1Lsolution}

Therefore there are 1.75 moles of BaCl_{2}

- For the Na_{2}SO_{4}

M=\frac{2.0moles[tex]Na_{2}SO_{4}}{1Lsolution}[/tex]

Therefore there are 2.0 moles of Na_{2}SO_{4}

2. Write the balanced chemical equation for the synthesis of the barium white pigment, BaSO_{4}:

BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the BaCl_{2}:

\frac{1.75}{1}=1.75

- For the Na_{2}SO_{4}:

\frac{2.0}{1}=2.0

As the BaCl_{2} is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = \frac{realyield}{theoreticalyield}*100

Percent yield = \frac{realyield}{408.45g}*100

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0
3 years ago
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