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Naddika [18.5K]
3 years ago
6

How do the valence electrons of an atom affect chemical reactions?

Chemistry
1 answer:
Inessa [10]3 years ago
3 0

Answer:

Electrons of an atom can affect chemical reactions, because electrons have a negative charge, which can introduce a form of energy.

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Write the full electron configuration of the Period 2 element with the following successive IEs (in kJ/mol):
Julli [10]

Answer:

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: 1s^22s^22p^1

Explanation:

Ionization Energy (IE):

It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.

If we look from left to right in a period, ionization energy increases due stability of valance shell.

From the data given to us:

IE₁ = 801

IE₂ = 2427

IE₃ = 3659

IE₄ = 25,022

IE₅ = 32,822

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: 1s^22s^22p^1

Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.

6 0
3 years ago
A valid lewis structure of __________ cannot be drawn without the central atom violating the octet rule. pf5 pcl3 so3 ccl4 co2
lakkis [162]
The answer will be, so3 since s only need two electrons to complete its octet
8 0
3 years ago
Please answer, I don't want to waste more points. ¿What's the answer to this Periodic Table question?
Pie

Answer:

sorry!!!

Explanation:

its to small I can't see it

5 0
3 years ago
If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of ai
Gnom [1K]

In accordance with Dalton's Law of multiple proportions

<h3>Further explanation</h3>

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

  • air(O₂) as a limiting reactant(product=14 g)

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

  • air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions

4 0
3 years ago
what is the temperature in Kelvin of 5.00 moles of nitrogen gas in a 30 L container with a pressure of 4.00 atm
Marianna [84]

The temperature of the nitrogen gas is 292.5 K.

<u>Explanation:</u>

Given that

Moles of Nitrogen, n = 5 mol

Volume, V = 30 L

Pressure, P = 4 atm

Gas Constant, R = 0.08205 L atm mol⁻¹ K⁻¹

Temperature = ? K

We have to use the ideal gas equation,

PV = nRT

by rearranging the equation, so that the equation becomes,

T = $\frac{PV}{nR}\\

Plugin the above values, we will get,

T = $\frac{4 \times 30}{5 \times 0.08205}

  = 292.5 K

So the temperature of the nitrogen gas is 292.5 K.

7 0
3 years ago
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