Explanation:
The given data is as follows.
= 98.70 kPa = 98700 Pa,
T =
= (30 + 273) K = 303 K
height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)
Density = 13.534 g/mL = 
= 13534 
The relation between pressure and atmospheric pressure is as follows.
P = 
Putting the given values into the above formula as follows.
P = 
= 
= 102683.05 Pa
= 102.68 kPa
thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.
The prefix milli means thousand so the correct conversion factor is 1000mg/g
Answer:
These particles were considered so small that they were unable to be cut
According to Avogadro's Law, same volume of any gas at standard temperature and pressure will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = ?
V = Volume = 16.8 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 16.8 L) ÷ 22.4 L
= 0.75 moles
Result:
16.8 L of Nitrogen gas will contain 0.75 moles at standard temperature and pressure.