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Vinil7 [7]
2 years ago
10

Q4 describe the change of atomic radio for the elements in period 2, from lithium to neon

Chemistry
1 answer:
mart [117]2 years ago
3 0

Answer:

Explanation:

Atomic radii of second period elements:

Li = [He] 2s¹   atomic radii = 123 pm

Be =  [He] 2s²  atomic radii = 111 pm

B =  [He] 2s² 2p¹   atomic radii = 86 pm

C =  [He] 2s² 2p²   atomic radii = 77 pm

N =  [He] 2s² 2p³   atomic radii = 74 pm

O =  [He] 2s² 2p⁴   atomic radii = 73 pm

F =  [He] 2s² 2p⁵   atomic radii = 72 pm

Ne = [He] 2s² 2p⁶   atomic radii = 154 pm

In case of neon weak van dar waals forces are present, there is less nuclear attraction that's why its atomic radii is greater.

Definition of atomic radii:

The atomic radius is the distance between center of two bonded atoms.

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.

Trend along group:

In group by addition of electron atomic radii increase from top to bottom due to increase in atomic number and addition of extra shell.

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A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te
Grace [21]

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

5 0
3 years ago
Which change can be easily be reversed? question 23 options: both a physical and chemical change chemical change physical change
givi [52]
Physical change .. you cannot undo chemical ... physical is like bending , shattering a mirror , denting something ..
chemical would be rust or something .

6 0
3 years ago
What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the
Verdich [7]

Answer:

[NaOH} = 0.4 M

Explanation:

In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.

(H₂SO₄, is considered strong, but the first deprotonation is weak)

2NaOH  +  H₂SO₄  →  Na₂SO₄  + 2H₂O

As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.

In the equivalence point we know mmoles of base = mmoles of acid

Let's finish the excersise with the formula

25 mL . M NaOH = 28.2 mL  .  0.355M

M NaOH = (28.2 mL  .  0.355M) / 25 mL → 0.400

8 0
3 years ago
An atom of the element iron has an atomic number of 26 and an atomic weight of 56. if it is neutral, how many protons, neutrons,
guapka [62]
Proton 26
neutron 0
you can tell how many proton by the atomic number
3 0
3 years ago
An element has a half-life of 30 years. if 1.0 mg of this element decays over a period of 90 years, how many mg of this element
labwork [276]

Answer:

144.6

Explanation:

6 0
3 years ago
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