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vaieri [72.5K]
3 years ago
14

A sub-shell with n = 6, l = 2 can accommodate a maximum of:

Chemistry
1 answer:
Mrrafil [7]3 years ago
8 0

Answer:

72

Explanation:

2n^2

n=6

2(6)^2=2×36

=72

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When 16.3 g of magnesium and 4.52 g of oxygen gas react, how many grams of magnesium oxide will be formed? Identify the limiting
Tanzania [10]

Answer:

22.77 g.

he limiting reactant is O₂, and the excess reactant is Mg.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:

no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.

no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.

So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.

<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>

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What is the difference between an atom and ion. What makes an ion an ion?​
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Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
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Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

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Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

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n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

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Now we have to calculate the mass of helium gas at doubled volume.

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Therefore, the mass of helium gas added must be 12.48 grams.

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Answer:

10/3 m/s. v=d/t. so 100/30 = 10/3

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