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Mariulka [41]
3 years ago
6

A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?

Chemistry
2 answers:
Artemon [7]3 years ago
7 0

Answer:

H₂O

Explanation:

The empirical formular of the compound is obtained using the following steps;

Step 1: Divide the percentage composition by the atomic mass

Hydrogen = 11.21 / 1 = 11.21

Oxygen = 88.79  / 16 = 5.55

Step 2: Divide by the lowest number

Hydrogen  = 11.21 / 5.55 = 2.02 ≈ 2

Oxygen = 5.55 / 5.55 = 1

This means the ratio of the elements is 2 : 1

The empirical formular (simplest formular of a compound) of the compound is;

H₂O

stiv31 [10]3 years ago
3 0

Answer:Empirical formula ======== H₂O    

Explanation:The empirical formula of a compound shows the whole number ratio for  each atom in a compound.

To find empirical formula. we follow the below steps

The total mass of the compound here  is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then  assume 11.21 grams  of hydrogen and 88.79grams of oxygen

                                          Hydrogen                   Oxygen

1.composition by mass    11.21                              88.79

molecular weight              1.007g/mol               15.990g/mol

2.Divide composition by mass  11.21/1.007            88.79/15.99    

by each molecular weight to get 11.13                            5.553

no of moles

3 Divide by the least number of moles

to get atomic ratio                       11.13/5.553          5.553/5.553

                                                         2.004                           1.00

4.Convert  to whole numbers             2                                 1

Empirical formula ======== H₂O    

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The enthalpy change for the reaction of titanium metal with gaseous iodine is given by the following thermochemical equation: 2
julia-pushkina [17]

Answer : The enthalpy change for the reaction is, 419.5 kJ

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According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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By reversing and then dividing the reaction by 2, we get the enthalpy change for the reaction.

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A compressed gas cylinder is filled with 5270 g of argon gas. The pressure inside the cylinder is 2050 psi at a temperature of 1
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Answer:

A compressed gas cylinder is filled with 5270 g of argon gas.

The pressure inside the cylinder is 2050 psi at a temperature of 18C.

The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.

How many grams of argon remains in the cylinder?

Explanation:

First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:

Mass of Ar gas is --- 5270g.

The number of moles of Ar gas:

Number of moles of Ar gas=\frac{given mass of Ar}{its atomic mass} \\                                             =\frac{5270g}{39.948g/mol} \\                                             =131.9mol

Temperature T=(18+273)K=291K

Pressure P=2050psi

2050* 0.0680atm\\\\=139.4atm\\

Volume V=?

PV=nRT\\139.4atm * V=131.9mol *0.0821L.atm.mol-1.K-1 * 291K\\=>V=\frac{131.9mol *0.0821L.atm.mol-1.K-1 * 291K}{139.4atm} \\=>V=22.6L

Using this volume V=22.6L

Pressure=650psi=44.2atm

Temperature T= (26+273)K=299K

calculate number of moles "n" value:

PV=nRT\\=>n=\frac{PV}{RT} \\=>n=\frac{44.2atm*22.6L}{0.0821L.atm.mol^-1K^-1* 299K} \\=>n=40.7mol

Mass of 40.7mol of Ar gas:

mass of Ar gas=number of moles * its atomic mass\\\\                        =40.7mol* 39.948g/mol\\\\                        =1625.8g

Answer:

The mass of Ar gas becomes 1625.8g.

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