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Artemon [7]
3 years ago
5

Need asap! Step-by-step pls: Determine the empirical formula: 32.37% Na, 22.58% S, 45.05% O​

Chemistry
1 answer:
Ludmilka [50]3 years ago
8 0

1) divide each percentage by the relative atomic mass of the element

2) divide all results by the smallest number

3)multiply by a whole number to get the simplest whole number ratio (if necessary)

that is to say:

Na                                   S                            O

32.37÷23                 22.58÷32                45.05÷16

= 1.407                       = 0.7056               = 2.816       (to 4 significant figures)

the smallest number here is 0.7056 so:

1.407÷0.7056        0.7056÷0.7056       2.816÷0.7056

=1.99 approx.2                   = 1                   3.99 approx. 4

here there is no need to carry out step 3 as ratio obtained is already a simplest whole number  ratio

so empirical formula is: Na₂SO₄

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<h3>What is Arrhenius acid-base reaction?</h3>
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To learn more about Arrhenius acid-base reaction: brainly.com/question/15196401

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4 0
1 year ago
A gas occupies 3.5L at 2.5 atm of pressure. What is the volume at 787 torr at the same temperature?
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Answer: 8.45 L

Explanation:

Given that,

Initial volume (V1) = 3.5L

Initial pressure (P1) = 2.5 atm

[Since final pressure is given in torr, convert 2.5 atm to torr

If 1 atm = 760 torr

2.5 atm = 2.5 x 760 = 1900 torr

Final volume (V2) = ?

Final pressure (P2) = 787 torr

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

3.5L x 1900 torr = 787 torr x V2

6650L•torr = 787 torr•V2

Divide both sides by 787 torr

6650L•torr/787 torr = 787 torr•V2/787 torr

8.45 L = V2

Thus, the volume of the gas at 787 torr and at the same temperature is 8.45 Liters

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3 years ago
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6 0
3 years ago
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Masteriza [31]
<h3><u>Answer; </u></h3>

=10.38  moles KOH  

<h3><u>Explanation</u>;</h3>

The balanced equation.  

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From the equation;

1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.  

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Moles of KOH;

1 mol Al2(SO4)3 : 6 mol KOH = 1.73  mol Al2(SO4)3 : x mol KOH  

Thus; x =  (6 × 1.73)

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4 0
3 years ago
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