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uysha [10]
2 years ago
6

How many paired and unpaired electrons are in magnesium?

Chemistry
1 answer:
pychu [463]2 years ago
8 0

Answer:

Paired = 12

Unpaired = 0

Explanation:

Magnesium is alkaline earth metal.

It is present in second group.

Its atomic number is 12 and atomic mass 24 amu.

Electronic configuration:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

It can seen from electronic configuration that all electrons are paired because s subshell have one orbital and it can accomodate two electrons. Each s subshells in magnesium have two electrons so these are filled and have paired electrons. While p subshell have three orbitals and can accomodate six electrons two by each orbital with opposite spin thus 2p is also filled and have paired electrons.

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A chemist is working on a reaction represented by this chemical equation:
erastova [34]

Answer is: A. 1.81 mol.

Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.

n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

6 0
2 years ago
Read 2 more answers
Oxalic acid is a diprotic acid. calculate the percent of oxalic acid (h2c2o4) in a solid given that a 0.7984-g sample of that so
vlada-n [284]
This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:

H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O

So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.

(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%

This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.
5 0
2 years ago
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A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0
2 years ago
Read 2 more answers
If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl
luda_lava [24]
I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by} = \frac{126580 g of benzyl chloride}{X}

Solving for X,
X = \frac{220 mg . 126580 mg}{353420 mg}
X = 78.79 mg

For PPh₃:
\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by} = \frac{262290 g of PPh3}{X}

Solving for X,
X = \frac{220 mg . 262290 mg}{353420 mg}
X = 163.27 mg

Now
, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when   \frac{95 percent}{100 percent} = \frac{220 g}{X}
Solving for X,

X = \frac{220 . 100}{95} = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when  \frac{220 mg phosphonium required}{231.57 mg require} = \frac{78.79 g of benzyl chloride}{X}
Solving for ,

X = \frac{231.57 . 78.79}{220} = 82.93 mg of Benzyl chloride

when  \frac{220 mg phosphonium required}{231.57 mg require} = \frac{163.27 g of PPh3}{X}
Solving for ,

X = \frac{231.57 . 163.27}{220} = 171.85 mg of PPh3

So,
reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.
7 0
2 years ago
What is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,
Karo-lina-s [1.5K]
The trick for this problem is to understand atomic mass: the fact that different atoms have different masses. What we need to do is add up all the atomic masses of the compound and work out the ratio of mass of water to the mass of sodium carbonate. Atomic masses are often given for each atom in the periodic table, but you can look them up on google too.

You can do this by adding up individual atoms for each molecule, or you can shortcut and lookup the molar mass of the compound (i.e.the task already done for you). 

The molar mass of water is 18.01g/mole so for 10 moles of water we have a mass of 180.1g.


The molar mass of sodium carbonate is 106g/mole (google).

So the total mass of the sodium carbonate decahydrate compound is 180.1+106 = 286.1g, of which water would make up 180.1g, so the percentage of water is is 180.1/286.1 = 0.629, so we can round this to 63%

:)




3 0
2 years ago
Read 2 more answers
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