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lianna [129]
3 years ago
12

The number of electrons in the outermost principal energy level of a chlorine atom is

Chemistry
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

seven electrons

Explanation:

Chlorine is present in group seventeen of periodic table. It is halogen element. All halogens have seven electrons in outer most valance shell.

The require only one electron to gain the stable electronic configuration or to complete the octet.

Electronic configuration of chlorine:

Cl₁₇ = 1s² 2s² 2p⁶ 3s² 3p⁵

Abbreviated electronic configuration:

Cl₁₇ = [Ne] 3s² 3p⁵

Properties of 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.

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A gas at 400 k is placed in a 4.0 l moveable piston. The gas is cooled to 252 k. What is the new volume of the piston?
katrin [286]

Answer:

2.52L

       

Explanation:

Given parameters:

T₁  = 400K

V₁  = 4L

T₂ = 252K

unknown

V₂ = ?

Solution:

To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.

Mathematically,

               \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

Substitute and solve for V₂

           \frac{4}{400}  = \frac{V_{2} }{252}

        V₂ = 2.52L

         

6 0
3 years ago
John is carefully pouring a chemical into a beaker when the beaker slips and breaks
Kryger [21]

Following laboratory safety protocols such as wearing personal protective equipment will protect John when the accident occurred.

<h3>What are laboratory safety protocols?</h3>

Laboratory safety protocols are the protocols put in place to ensure safety in the laboratory.

Laboratory safety protocols include the following:

  • always wear personal protective equipment in the laboratory
  • do not play in the laboratory
  • do not eat in the laboratory

Following laboratory safety protocols will help protect us from accidents which occur in the laboratory.

What happened when john was carefully pouring a chemical into a beaker when the beaker slips and breaks is an example of laboratory accident.

Wearing personal protective equipment will protect John.

In conclusion, following laboratory safety protocols will protect us when accidents occur in the laboratory.

Learn more about laboratory safety protocols at: brainly.com/question/17994387

#SPJ1

Note that the complete question is given as follows:

John is carefully pouring a chemical into a beaker when the beaker slips and breaks. How would laboratory safety protocols help John?

5 0
2 years ago
How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +
krek1111 [17]

Answer:  13.9 g of H_2O will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles

The balanced chemical reaction is:

4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)

According to stoichiometry :

7 moles of O_2 produce =  6 moles of H_2O

Thus 0.900 moles of O_2 will produce =\frac{6}{7}\times 0.900=0.771moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g

Thus 13.9 g of H_2O will be produced from the given mass of oxygen

5 0
3 years ago
The amount of matter in an object is known as
zavuch27 [327]
Mass......................

4 0
3 years ago
Read 2 more answers
12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?
bazaltina [42]

Answer:

u=4.57x10^5GHz

Explanation:

Hello.

In this case, given the formula:

E=h*u

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\  \\u=4.57x10^{14}s^{-1}

Or also:

u=4.57x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ \\u=4.57x10^5GHz

Best regards.

5 0
3 years ago
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