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Umnica [9.8K]
3 years ago
11

A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?

Chemistry
1 answer:
Sergio039 [100]3 years ago
7 0

We will determine the moles of each element first

moles = mass / atomic mass

Atomic weight of Al : 27 g/ mol

Atomic weight of P : 3 1g /mol

Atomic weight of O : 16 g /mol

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g


Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40/ 31 = 0.819

Moles of O = mass / atomic mass = 52.50/ 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

moles of Al = 0.819 / 0.819 = 1

moles of P = 0.819 / 0.819 = 1

moles of O = 3.28 / 0.819 =  4

So the empirical formula will be  : AlPO4


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Answer:

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The Ionic Equation is ...

NH₄⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) => NH₄⁺(aq) + NO₃⁻(aq) + AgCl(s)

This shows all ions from reaction plus the Driving Force of the reaction.

The Net Ionic Equation is ...

Ag⁺(aq) + Cl⁻(aq) => AgCl(s)

The Net Ionic Equation shows only those ions undergoing reaction. The NH₄⁺ and NO₃⁻ ions are 'Spectator Ions' and do not react.

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4 0
3 years ago
7. Convert 5.2 cm of magnesium (Mg) ribbon to mm of Mg ribbon.
Effectus [21]
<h3>Answer:</h3>

52 mm

<h3>Explanation:</h3>

We are given;

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Required to convert it to cm

We are going to use the appropriate conversion factor;

  • The units used to measure length include;

Kilometer(km)

10

Hectometer (Hm)

10

Decameter (dkm)

10

Meter(m)

10

Decimeter (dm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore; the appropriate conversion factor is 10mm/cm

Thus;

5.2 cm will be equivalent to;

= 5.2 cm × 10 mm/cm

= 52 mm

Therefore, the length of magnesium ribbon is 52 mm

7 0
2 years ago
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given that 32.0g sulphur contains 6.02×10^23 sulphur atoms,how many atoms are there in 2.70g of aluminum​
dangina [55]

Answer:

6.022 × 10²² atoms

Explanation:

Generally 1 mol of any element contains 6.02×10^23  atoms. The number 6.022 × 10²³ is known as Avogadro's number.

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Molar mass = 27g/mol

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0.1 mol = x

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Answer:

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