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Natalija [7]
3 years ago
11

How many atoms of fluorine are present in a molecule of carbon tetrafluoride, cf4?

Chemistry
2 answers:
Stella [2.4K]3 years ago
8 0

Answer : The number of atoms of fluorine are present in a molecule of carbon tetrafluoride are, 4 atoms.

Explanation :

Molecules : It is defined as the group of atoms which are bonded together and they represent the smallest fundamental unit of chemical compound.

It can be molecules of element (combination of same kind of atoms) or molecules of compound (combination of different kind of atoms).

The given molecule is carbon tetrafluoride and the chemical formula of carbon tetrafluoride is, CF_4

In carbon tetrafluoride, there are 1 number of carbon atom and 4 number of fluorine atoms.

Hence, the number of atoms of fluorine are present in a molecule of carbon tetrafluoride are, 4 atoms.

Ivanshal [37]3 years ago
3 0
There are four atoms of fluorine present
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Which particle in an atom is lost or gained to make an ion
denpristay [2]

An electron because that is the only part able to be lost or gained without nuclear action needed

7 0
3 years ago
Covert 1.57 x 10^14 nm= ___m
LekaFEV [45]

Answer:

1.57 x 10⁷m

Explanation:

Given quantity is;

          1.57 x 10¹⁴nm

Now;

            1 nm  = 10⁻⁹

So, let us convert this given quantity;

         1 nm          =             10⁻⁹

  1.57 x 10¹⁴nm  will give 1.57 x 10¹⁴ x 10⁻⁹  = 1.57 x 10⁷m

 

7 0
3 years ago
If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate
NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0
3 years ago
How many grams of aluminum are needed to completely react with<br> 192 g of oxygen gas?
OleMash [197]

Answer:

4.28g

Explanation:

6 0
3 years ago
What is the common name of the ether that has the following structure? H3C----O-----CH2CH3
fomenos

Answer:

Ethyl methyl ether.

Explanation:

  • The mono ether is named commonly by <em>dividing the molecule into two parts, writing them in alphabetical order, and adding ether in the end of the name.</em>
  • In this compound; we have two parts CH₃ (methyl) and CH₃CH₂ (ethyl).

So, it is named ethyl methyl ether.

  • The IUPAC name of this name is Methoxyethane.
4 0
4 years ago
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