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Alisiya [41]
3 years ago
9

What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorin

e and the rest is bromine
Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

  • 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:

  • 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
  • 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
  • 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we <u>divide those numbers of moles by the lowest number among them</u>:

  • 0.9594 mol Mo / 0.9594 = 1
  • 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
  • 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

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Answer:

                      %H  = 6.72 %

Explanation:

                   Percent composition of an element is the total mass of that element divided by the molecular mass of compound (or molecular mass) of which it is present in.

So,

Percent composition of Hydrogen will be given as,

         %H  =  Total mass of H / Molecular Mass of Acetic Acid × 100

So,

Total Mass of H  =  1.01 × 4 = 4.04 g

Molecular Mass of Acetic acid  =  60.052 g/mol

Putting values in above formula,

         %H  =  4.04 g/mol ÷ 60.052 g/mol × 100

         %H  = 6.72 %

6 0
2 years ago
In the bromination of arenes, which of the following statements regarding the reaction is true?a.The hydrocarbon is used in exce
Olenka [21]

The hydrocarbon is used in excess.

<h3><u>Explanation</u>:</h3>

The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.

8 0
3 years ago
The green areas on the diagram show the world's rainforests. What can you conclude about the location of rainforests?
Ber [7]

Answer:

Option A

They are located near the equator

Explanation:

From the diagram, we can observe that all the rain forest zones occur near the equator.

This is because on an annual basis, the equator receives the highest concentration of solar radiation in the whole Earth. This implies that the rate of evaporation of water from the water bodies present  is high, and consequently, the amount of rainfall is high also. This abundant sunshine and high rainfall  leads to the growth of the tropical rain forests at those regions

6 0
2 years ago
Calculate average kinetic energy of one mole of gas at 517K
Gekata [30.6K]

In an ideal gas, there are no attractive forces between the gas molecules, and there is no rotation or vibration within the molecules. The kinetic energy of the translational motion of an ideal gas depends on its temperature. The formula for the kinetic energy of a gas defines the average kinetic energy per molecule. The kinetic energy is measured in Joules (J), and the temperature is measured in Kelvin (K).

K = average kinetic energy per molecule of gas (J)

kB = Boltzmann's constant ()

T = temperature (k)

Kinetic Energy of Gas Formula Questions:

1) Standard Temperature is defined to be . What is the average translational kinetic energy of a single molecule of an ideal gas at Standard Temperature?

Answer: The average translational kinetic energy of a molecule of an ideal gas can be found using the formula:

The average translational kinetic energy of a single molecule of an ideal gas is  (Joules).

2) One mole (mol) of any substance consists of  molecules (Avogadro's number). What is the translational kinetic energy of  of an ideal gas at ?

Answer: The translational kinetic energy of  of an ideal gas can be found by multiplying the formula for the average translational kinetic energy by the number of molecules in the sample. The number of molecules is  times Avogadro's number:

7 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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