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adelina 88 [10]
3 years ago
6

How many fluorine atoms are present in 5.85 g of c2f4?

Chemistry
1 answer:
kozerog [31]3 years ago
4 0
The molar mass of C2F4 (tetrafluoroethylene) is equal to 100 g/mol. The number of moles of C2F4 in the given amount,
                               n = (5.85 g) / (100 g/mol) = 0.0585 mols C2F4
The number of molecules per mol is calculated through the equation,
                          (0.0585 mols) x (6.022 x 10^23)
Since there are 4 F atoms each molecule then, 
                           (0.0585 mols) x (6.022 x 10^23)(4)
                          = 1.41 x 10^23 atoms of F
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PLEASE HELP
neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

               = 144 + 22 + 176

               = 342 g

2.- Convert the mass of sucrose to moles

                  342 g of sucrose ------------------- 1 mol

                  125 g of sucrose -------------------- x

                           x = (125 x 1) / 342

                           x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0
3 years ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
The alkali metals play virtually the same general chemical role in all their reactions. (b) How is it based on atomic properties
vfiekz [6]

b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.

Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.

Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:

  • Lithium (Li)
  • Sodium (Na)
  • Potassium (K)
  • Rubidium (Rb)
  • Cesium (Cs)
  • Francium (Fr)

Properties of alkali metals are:  Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.

To know more about Alkali metals, refer to this link:

brainly.com/question/18153051

#SPJ4

6 0
1 year ago
In what situation do we use a volumetric flask, conical flask, pipette and graduated cylinder? Explain your answer from accuracy
Ad libitum [116K]
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring. 
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
3 0
3 years ago
The concentration of a salt solution in terms of mass by mass percentage is 20% and the mass of the solution is 500g determine t
marin [14]
100 grams of solute would be present as 20% of 500 is 100
5 0
3 years ago
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