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3 years ago

Please answer asap need it by Wednesday morning

Chemistry

1 answer:

yarga [219]3 years ago

4 0

The answer to this problem is quite simple, it’s 9

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The elements from this section of the periodic table all belong to the same

krok68 [10]

Answer:

Option C = same period.

Explanation:

All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell.

However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.

7 0

3 years ago

Read 2 more answers

Homework . Answered

bija089 [108]

We have that the name (not chemical symbol) of the main group element in period 5 and group 3A is

Metalloid boron (B)

From the Question we are told that

The element belongs to

Period 5 and group 3A

Generally

Group 3A of the periodic table includes the metalloid boron (B), aluminum (Al), indium (In), and thallium (Tl) gallium (Ga),

Period 5 is possessed by the metalloid boron (B) of the Group 3A

For more information on this visit

brainly.com/question/13025901?referrer=searchResults

7 0

3 years ago

How many moles of h2o are produced in neutralization when 1. 5 mole of mg(oh)2 reacts with h2so4?

Firlakuza [10]

There are 3 moles of $H_{2} O$ are produced in neutralization when 1. 5 mole of $Mg(OH)_{2}$ reacts with $H_{2} SO_{4$

The neutralization reaction between magnesium hydroxide and sulfuric acid is given as,

$Mg(OH)_{2} + H_{2} SO_{4$ → $MgSO_{4} + 2H_{2} O$

from above chemical equation , we can conclude the molar ratio = 1:2

It means two moles of water molecule is produce from one mole of magnesium hydroxide.

So, number of moles of water molecules produced from 1. 5 mole of magnesium hydroxide = 1. 5 mole× 2 = 3 moles

Therefore, number of moles of water molecules produced from 1. 5 mole of magnesium hydroxide is 3 moles.

To learn more about neutralization

brainly.com/question/27891712

#SPJ4

5 0

2 years ago

Is Tetracarbon decoxide ionic or covalent?

Cloud [144]

Answer: covalent bond

Explanation: Tetracarbon Hexahyride is built using covalent bonds

Hope this helps please mark brainlist it would be greatly appreiciated :)

8 0

3 years ago

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water

maks197457 [2]

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}\times \text{Volume of ethylene glycol}=1.114g/mL\times 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/mL\times 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{2.114g}{1.070g/mL}=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}\times 100=\frac{1mL}{1.975mL}\times 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}\times 100=\frac{1.114g}{2.114g}\times 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Volume of solution (in mL)}}$

$\text{Molarity}=\frac{1.114g\times 1000}{62.07g/mole\times 1.975L}=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water (in g)}}$

$\text{Molality}=\frac{1.114g\times 1000}{62.07g/mole\times 1kg}=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=\frac{1.114g}{62.07g/mole}=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=\frac{1g}{18g/mole}=0.0555mole$

$\text{Mole fraction of ethylene glycol}=\frac{0.01795mole}{0.01795mole+0.0555mole}=0.244$

6 0

3 years ago