Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
<span>The molar mass of the compound is 122 g. </span>
Answer:


Explanation:
Hello,
In this case, we can compute the mole fraction of benzene by using the following formula:

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

Thus, we compute the mole fraction:

Next, for the molality, we define it as:

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

Or just 0.990 m in molal units (mol/kg).
Best regards.
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O
From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.
Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=100 kPa
T=293 K
R=8.314472 L kPa K⁻¹ mol⁻¹
Number of moles of gas=3.43 mole
Putting all the values in the above equation,

V=83.55 L
So the volume will be 83.55 L.
83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).