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Colt1911 [192]
3 years ago
8

AICI, + NaNaCl + Al Did Al change oxidation number? yes No

Chemistry
1 answer:
mezya [45]3 years ago
3 0

<u>Answer:</u>

<em>Yes, Al change oxidation number</em>

<em></em>

<u>Explanation:</u>

AlCl_3  +Na >NaCl+Al

For a free element, oxidation number is 0.

Na and Al are free elements we find in the equation as they are not in combination with other elements.

For a compound, the sum of oxidation number of elements = 0.

We know oxidation number of chlorine is -1.

So let us find the oxidation number of Al in AlCl_3

Let x be the oxidation number of Al so,

AlCl_3=x+(3\times-1)=0\\\\x-3=0\\\\ x= +3

Oxidation number of Al in AlCl_3  is +3

Oxidation number of Al  changes from +3 to 0.

Decrease in oxidation number indicates AlCl_3 is getting reduced in the reaction.

Na oxidation number increases from 0 to +1 shows Na is getting oxidised in the reaction  

The change in oxidation number of the elements represents a redox reaction where both oxidation and reduction takes place simultaneously.

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How many molecules are in 3.50 moles of H2O<br>​
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Answer:

2.11 x 10²⁴ molecules.

Explanation:

  • <em>It is known that every 1.0 mole of a molecule contains Avogadro's number of molecules (NA = 6.022 x 10²³).</em>

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

1.0 mole of H₂O contains → 6.022 x 10²³ molecules.

3.5 mole of H₂O contains → ??? molecules.

∴ 3.5 mole of H₂O contain = (3.5 mol)(6.022 x 10²³) = 2.11 x 10²⁴ molecules.

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How many atoms are in the following: 3Ca3(PO4)2?<br> A 3<br> B)9<br> С)13<br> D) 39
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5 0
3 years ago
0.265g of an organic compound produced an evaporation 102cm³ of vapour at 373k and 775mmHg percentage composition of the constit
kumpel [21]

A. The molecular mass of the compound is 77.9 g/mol

B. The molecular formula of the compound is C₆H₆

<h3><u>Determination of the mole of the compound</u></h3>

We'll begin by calculating the number of mole of compound using the ideal gas equation as shown below:

  • Volume (V) = 102 cm³ = 102 / 1000 = 0.102 L
  • Temperature (T) = 373 K
  • Pressure (P) = 775 mmHg = 775 / 760 = 1.02 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

n = PV / RT

n = (1.02 × 0.102) / (0.0821 × 373)

n = 0.0034 mole

<h3><u>A.</u><u> Determination of the </u><u>molecular mass</u><u> of the </u><u>compound</u><u>. </u></h3>
  • Mass = 0.265 g
  • Number of mole = 0.0034 mole
  • Molecular mass =?

Molecular mass = mass / mole

Molecular mass = 0.265 / 0.0034

Molecular mass of compound = 77.9 g/mol

<h3><u>B</u><u>. Determination of the </u><u>molecular formula</u><u> of the compound. </u></h3>

We'll begin by calculating the empirical formula of the compound.

  • Carbon (C) = 92.24%
  • Hydrogen (H) = 7.76%

Empirical formula =?

Divide by their molar mass

C = 92.24 / 12 = 7.69

H = 7.76 / 1 = 7.76

Divide by the smallest

C = 7.69 / 7.69 = 1

H = 7.76 / 7.69 = 1

Thus the empirical formula of the compound is CH

Finally, we shall determine the molecular formula.

  • Molecular mass = 77.9 g/mol
  • Empirical formula = CH
  • Molecular formula =?

Molecular formula = empirical × n = molecular mass

[CH]n = 77.9

[12 + 1]n = 77.9

13n = 77.9

Divide both side by 13

n = 77.9 / 13

n = 6

Molecular formula = [CH]n

Molecular formula = [CH]₆

Molecular formula = C₆H₆

Complete Question:

0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the constituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition.

Learn more about ideal gas equation:

brainly.com/question/14364992

Learn more about molecular formular:

brainly.com/question/512891

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