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UNO [17]
3 years ago
13

A compound having an approximate molar mass of 165 - 170 g has the following...

Chemistry
1 answer:
VladimirAG [237]3 years ago
3 0

Answer:

Find ratio of atoms for empirical formula, divide mass by atomic mass

C=42.87/12 = 3.57

H= 3.598/1 = 3.598

O= 28.55/16 = 1.784

N= 25.00/14 =1.1785

to see ratio, divide by smallest no

Gives C2H2ON for empirical formula

Empirical mass = 24 +2 +16 + 14 = 56

165 to 170 /56 = 3

so molecular formula = empirical formula x 3= C6H6O3N3

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MissTica
<span>3) P4O10 and P2O5
The mass of the P4O10 divided by the mass of the P2O5=2, and if you multiply the number of atoms in the P2O5 by 2, you get the P4O10, thus P2O5 is its empirical formula.</span>
3 0
4 years ago
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How many moles of sodium carbonate contain 1.773 × 1017 carbon atoms?
vesna_86 [32]

Answer:

  • 2.941 × 10⁻¹⁷ mol

Explanation:

1) Chemical formula of sodium carbonate: <em>Na₂CO₃</em>

2) Ratio of carbon atoms:

  • The number of atoms of C in the unit formula Na₂CO₃ is the subscript for the atom, which is 1 (since it is not written).

Hence, the ratio is 1 C atom / 1 Na₂CO₃ unit formula.

This is, there is 1 atom of carbon per each unit formula of sodium carbonate.

3) Calculate the number of moles in 1.773 × 10⁷ carbon atoms

  • Divide by Avogadro's number: 6.022 × 10²³ atoms / mol

  • number C moles  = 1.773 × 10⁷ atoms / (6.022 × 10²³ atoms/mol)

  • number C moles = 2.941 × 10⁻¹⁷ mol

Since, the ratio is 1: 1, the number of moles of sodium carbonate is the same number of moles of carbon atoms.

5 0
3 years ago
The term “periodic” helps identify _______ that occur in the table, like electronegativity or atomic radii.
stepan [7]

Answer: ITS C OR B I THINK BUT I HAVE A FEELING ITS ONE OF THOSE

Explanation:

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Explanation:

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Answer:

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Explanation:

The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals

It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.

Its atomic number is 3. The electronic configuration is 1s2 2S1

5 0
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