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PIT_PIT [208]
3 years ago
12

Sodium electron configuration

Chemistry
1 answer:
Alex73 [517]3 years ago
4 0

Full:

1s² 2s² 2p⁶ 3s¹

Abbreviated:

[Ne] 3s¹

You might be interested in
Balance the equation for magnasuin and oxygen to yield magnasuimoxide
Minchanka [31]

Answer:

2Mg + O₂  →  2MgO

Explanation:

Chemical equation:

Mg + O₂  →  MgO

Balanced chemical equation:

2Mg + O₂  →  2MgO

The balanced equation s given above and it completely follow the law of conservation of mass.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Steps to balanced the equation:

Step 1:

Mg + O₂  →  MgO

Mg = 1         Mg = 1

O = 2           O = 1

Step 2:

2Mg + O₂  →  MgO

Mg = 2         Mg = 1

O = 2           O = 1

Step 3:

2Mg + O₂  →  2MgO

Mg = 2         Mg = 2

O = 2           O = 2

6 0
3 years ago
Anhydrous copper sulphate can be used in a test for water. What two things will happen when water is added to anhydrous copper s
Misha Larkins [42]

Answer:

thanks bro for helping me thanku so much

3 0
2 years ago
An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?
aleksley [76]

Answer:- Formula of the hydrate is Fe_2(SO_4)_3.5H_2O and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})

= 0.204molFe_2(SO_4)_3

18.4gH_2O(\frac{1mol}{18.02g})

= 1.02molH_2O

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

Fe_2(SO_4)_3=\frac{0.204}{0.204}  = 1

H_2O=\frac{1.02}{0.204} = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is Fe_2(SO_4)_3.5H_2O and the name of the hydrate is Iron(III)sulfate pentahydrate.


3 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
3 years ago
Read 2 more answers
An object has a mass of 20g and occupies a volume of 40 ml. Calculate the Density of the object.
Vsevolod [243]

Answer:

<h3>The answer is 0.5 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{20}{40}  \\

We have the final answer as

<h3>0.5 g/mL</h3>

Hope this helps you

3 0
3 years ago
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