Answer:
Third Law of motion. (answer C)
Explanation:
This is an example of Newton's Third Law of motion which normally translates into Action/Reaction. That is : the expelled air exerts a force in the direction opposite to that of the motion of the air that goes out. It is the same as propulsion in a rocket.
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Explanation:
<u>Moles is denoted by given mass divided by the molecular mass , </u>
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
For example ,
For a compound X ,
The given mass i.e. w = 20 g
and the molecular mass ,i.e. , m = 10 g / mol
Then the moles can easily be calculated by using the above formula ,
n = w / m
n = 20 g / 10 g/mol = 2 mol
Hence , answer = 2 mol.
Answer:
Volume of container = 0.0012 m³ or 1.2 L or 1200 ml
Explanation:
Volume of butane = 5.0 ml
density = 0.60 g/ml
Room temperature (T) = 293.15 K
Normal pressure (P) = 1 atm = 101,325 pa
Ideal gas constant (R) = 8.3145 J/mole.K)
volume of container V = ?
Solution
To find out the volume of container we use ideal gas equation
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First we find out number of moles
<em>As Mass = density × volume</em>
mass of butane = 0.60 g/ml ×5.0 ml
mass of butane = 3 g
now find out number of moles (n)
n = mass / molar mass
n = 3 g / 58.12 g/mol
n = 0.05 mol
Now put all values in ideal gas equation
<em>PV = nRt</em>
<em>V = nRT/P</em>
V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa
V = 121.87 ÷ 101,325 pa
V = 0.0012 m³ OR 1.2 L OR 1200 ml