The answer to this problem is 11.6m
Answer:
A more dense plate going underneath a less dense plate.
Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,

<u>0.7457 g is the mass of the helium gas. </u>
The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.
Molar mass of the unknown compound = 223.94 g/mol (given)
Mass of each element present in the unknown compound is determined as:
- Mass of carbon,
:

- Mass of hydrogen,
:

- Mass of chlorine,
:

Now, the number of each element in the unknown compound is determined by the formula:

- Number of moles of
:

- Number of moles of
:

- Number of moles of


Dividing each mole with the smallest number of mole, to determine the empirical formula:


Multiplying with 2 to convert the numbers in formula into a whole number:
So, the empirical formula is
.
Empirical mass = 
In order to determine the molecular formula:
n = 
n = 
So, the molecular formula is:
