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lutik1710 [3]
3 years ago
13

As atoms get smaller,valance electrons are getting closer or farther away?

Chemistry
2 answers:
lutik1710 [3]3 years ago
8 0

Answer:

valence electrons get closer but rest of them are remain same

Fed [463]3 years ago
7 0
They are getting closer and the force of attraction is getting weaker
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An increase in the atomic number________the atomic radius moving from left to right across a period.
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Answer:

decrease

Explanation:

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It is the distance from the center of nucleus to the outer most electronic shell.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required

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How many moles are in 4.5 g of<br> Sodium Chloride, NaCl?
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Which sample of water contains particles having the highest kinetic energy
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1. A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C. If the
Maurinko [17]

Answer:

Mass of aluminium in sample = 3.591 g ≅ 3.6 grams

Explanation:

Given that,  A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.

the specific heat of aluminum is 0.900 J/g- °C

The relation between heat absorbed and change in temperature is given by,   Q = msΔT.

where Q = heat absorbed

            m = mass of the substance

            s = specific heat of substance

          ΔT  = change in temperature

Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C

⇒ m =  \frac{Q}{s(T_{2} -T_{1}) }

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⇒ m ≅ 3.6 g

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3 years ago
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