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slava [35]
3 years ago
6

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

2 were formed. what is the simplest formula of the compound?
Chemistry
1 answer:
SVEN [57.7K]3 years ago
3 0

Answer:- CH_2O_2

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

1.17gH_2O(\frac{1mol}{18.02g})

= 0.0649molH_2O

Similarly, 2.87gCO_2(\frac{1mol}{44g})

= 0.0652molCO_2

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

2.0868gO(\frac{1mol}{15.999g})

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = \frac{0.0652}{0.0652}  = 1

H = \frac{0.1298}{0.0652}  = 2

O = \frac{0.130}{0.0652}  = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is CH_2O_2 .



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6 0
3 years ago
Rank the following elements by effective nuclear charge, Zeff, for a valence electron. F LI Be B N
Stels [109]

Answer:

Rank in increasing order of effective nuclear charge:

  • Li < Be < B < N < F

Explanation:

This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.

<u>1) Effective nuclear charge definitions</u>

  • While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

  • Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

<u><em>2) Z eff for a F valence electron:</em></u>

  • F's atomic number: Z = 9
  • Total number of electrons: 9 (same numer of protons)
  • Period: 17 (search in the periodic table or do the electron configuration)
  • Number of valence electrons:  7 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
  • Zeff = Z - S = 9 - 2 = 7

<u><em>3) Z eff for a Li valence eletron:</em></u>

  • Li's atomic number: Z = 3
  • Total number of electrons: 3 (same number of protons)
  • Period: 1 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 1 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
  • Z eff = Z - S = 3 - 2 = 1.

<em>4) Z eff for a Be valence eletron:</em>

  • Be's atomic number: Z = 4
  • Total number of electrons: 4 (same number of protons)
  • Period: 2 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 2 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
  • Z eff = Z - S = 4 - 2 = 2

<u><em>5) Z eff for a B valence eletron:</em></u>

  • B's atomic number: Z = 5
  • Total number of electrons: 5 (same number of protons)
  • Period: 13 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 3 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
  • Z eff = Z - S = 5 - 2 = 3

<u><em>6) Z eff for a N valence eletron:</em></u>

  • N's atomic number: Z = 7
  • Total number of electrons: 7 (same number of protons)
  • Period: 15 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 5 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
  • Z eff = Z - S = 7 - 2 = 5

<u><em>7) Summary (order):</em></u>

  Atom          Zeff for a valence electron

  • F                   7
  • Li                   1
  • Be                 2
  • B                   3
  • N                   5

  • <u>Conclusion</u>: the order is Li < Be < B < N < F
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Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,,

Volume =\frac{Moles\ of\ solute}{Molarity}

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

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3 years ago
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3 years ago
How many grams of sodium are needed to produce 12.5g of sodium oxide
Hatshy [7]

Answer:

25 possibly

Explanation:

I'm not too sure about this, but sodium oxide is Na2O, 2 sodium and 1 oxygen, so 12.5g * 2 is 25

If someone else comes up with a more convincing argument listen to them

4 0
3 years ago
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