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3 years ago

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What quantum numbers specify a 5p orbital?

1 answer:

mezya [45]3 years ago

8 0

n = 5

l = 0,1,2,3,4

ml = -4,-3,-2,-1,0,1,2,3,4,

ms = +1/2 and -1/2

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A 0.100-l sample of an unknown hno3 solution required 32.9 ml of 0.200 m ba(oh)2 for complete neutralization. what was the conce

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The reaction between HNO3 and Ba(OH)2 is given by the equation below;
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
Moles of Barium hydroxide used;
= 0.200 × 0.039 l
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The mole ratio of HNO3 and Ba(OH)2 is 2: 1
Therefore; moles of nitric acid used will be;
= 0.0078 ×2 = 0.0156 moles
But; 0.0156 moles are equal to a volume of 0.10
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If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are p

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<u>Answer:</u> The mass of iron (III) oxide produced is 782.5 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For </u> $FeS_2$ <u> :</u>

Given mass of $FeS_2$ = 588 g

Molar mass of $FeS_2$ = 120 g/mol

Putting values in equation 1, we get:

$\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol$

<u>For </u> $O_2$ <u> :</u>

Given mass of $O_2$ = 352 g

Molar mass of $O_2$ = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol$

The chemical equation for the reaction of $FeS_2$ and oxygen gas follows:

$FeS_2+O_2\rightarrow Fe_2O_3+SO_2$

By Stoichiometry of the reaction:

1 mole of $FeS_2$ reacts with 1 mole of oxygen gas

So, 4.9 moles of $FeS_2$ will react with = $\frac{1}{1}\times 4.9=4.9mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, $FeS_2$ is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of $FeS_2$ produces 1 mole of iron (III) oxide

So, 4.9 moles of $FeS_2$ will produce = $\frac{1}{1}\times 4.9=4.9moles$ of iron (III) oxide

Now, calculating the mass of iron (III) oxide from equation 1, we get:

Molar mass of iron (III) oxide = 159.7 g/mol

Moles of iron (III) oxide = 4.9 moles

Putting values in equation 1, we get:

$4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g$

Hence, the mass of iron (III) oxide produced is 782.5 grams

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3 years ago