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SSSSS [86.1K]
3 years ago
10

What is the ground-state electron configuration of chlorine and the number of valence electrons in it?

Chemistry
1 answer:
SSSSS [86.1K]3 years ago
8 0

Explanation:

Chlorine is an element in group 7 on the periodic table.

  It is the second element in this group of halogens

 The element has a total of 17 electrons:

Ground state configuration  = 1s² 2s² 2p⁶ 3s² 3p⁵

Valence electrons are the outermost shell electrons;

  From the configuration: 3s² 3p⁵ is the valence shell

           Number of electrons  = 2 + 5  = 7

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3 years ago
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How many moles are in 74.8 grams of NaCl?
Pie

Answer:

1.28 mol NaCl

Explanation:

Find the mass of NaCl by adding the two elements atomic mass

Na: 22.99

Cl: 35.45

22.99+35.45=58.44 g

Now convert to moles

74.8 g / 58.44 g = 1.27994524298 mol

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3 years ago
A swimming pool has the dimensions of 10m, 5m, and 1m. what is the volume?
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The diagram above shows the repeating groups of atoms that make
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3 years ago
Be sure to answer all parts. A person drinks four glasses of cold water (3.2 degree C) every day. The volume of each glass is 2.
Lelechka [254]

Explanation:

(a). The given data is as follows.

Volume of one glass of water = 2.2 \times 10^{2} ml = 220 ml

Volume of 4 glass of water = 220 \times 4 = 880 ml

We known that density of water is 1 g/ml.  Therefore, calculate the mass of water as follows.

       Mass of water = 880 ml \times 1 g/ml

                               = 880 gm

                               = 0.88 Kg               (as 1 kg = 1000 g)

The relation between heat energy, mass and temperature change is as follows.

            Q = mC \Delta T

\Delta T = (37 - 3.2)^{o}C = 33.8^{o}C

Putting the given values into the above formula as follows.

         Q = mC \Delta T

             = 0.88 \times 4.186 J/g^{o}C \times 33.8^{o}C

             = 124.5 kJ

Hence, the body have to supply 124.5 kJ to raise the temperature of the water to 37 degree C.

(b).      As we know that the heat of fusion of ice is 333 J/g.

So, energy required for 8.4 \times 10^{2} g or 840 g is as follows.

          333 \times 840 = 279.72 kJ

Heat capacity of water= 4.184 J/g^{o}C

Now, heat energy will be as follows.

           Q = 4.184 \times 840 g \times 37^{o}C

               = 130.03 kJ

Therefore, total heat required = (279.72 + 130.03) kJ

                                                  = 409.75 kJ

Hence, for the given situation your body should lose 409.75 kJ  of heat.

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