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Vika [28.1K]
3 years ago
9

According to the Pauli exclusion principle, when can two electrons occupy the same orbital?

Chemistry
1 answer:
Lyrx [107]3 years ago
3 0
C.

For electrons to occupy the same orbital, every other orbital in the sub shell must have at least one electron and the paired electrons must have opposite spins.
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An aqueous solution of sodium
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Answer:

2.5×10^-7mol/dm^3

Explanation:

Firstly convert the cm^3 to dm^3

200×1000=200000dm^3

Calculate the g/dm^3

2/200000=0.00001g/dm^3

To calculate mol/dm^3

Mol/dm^3=mass given\molar mass

=0.00001/40

=2.5×10^-7mol/dm^3

8 0
3 years ago
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A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemic
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3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:
slamgirl [31]

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

8 0
3 years ago
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in this punnett square what is the ratio of homozgous to heterozygous offspring? in the punnett square what percentage of the of
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Idk if this is a ratio but I know the percentage will be 3/4 Tt and 1/4 tt so Tt is 75% tt is 25%
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