Answer:
Partial pressure of oxygen = 23.38 kpa (Approx)
Explanation:
Given:
Amount of oxygen = 23.15%
Amount of nitrogen = 76.85%
Pressure (missing) = 101 kpa
Find:
Partial pressure of oxygen
Computation:
Partial pressure of oxygen = [Amount of oxygen x Pressure]/100
Partial pressure of oxygen = [23.15% x 101]/100
Partial pressure of oxygen = 23.38 kpa (Approx)
Answer:
The correct answer is option c.
Explanation:
Formula used to determine an average atomic mass :
Mass of isotope Sb-121 = 120.904 amu
Fractional abundance of Sb-121 = 57.21% = 0.5721
Mass of isotope Sb-123 = 122.904 amu
Fractional abundance of Sb-123 = 42.79% = 0.4279
Average atomic mass of Sb:
You HAVE TO know that molarity (M) tells you the number of moles of a solute per Liters of solution.
M=mol/L
0.3M = 2.7 mol/ Volume
Volume = 2.7 mol/3.0 L
Volume = 0.9 L
The mass will stay the same because of the conservation of mass
<h3>
Answer:</h3>
5.6 Liters
<h3>
Explanation:</h3>
- N.T.P. refers to the standard temperature and pressure (S.T.P).
We need to know that;
- One mole of a gas occupies a volume of 22.4 liters at N.T.P.
In this case;
We have 11 g of CO₂
But, 1 mole of CO₂ occupies 22.4 l at N.T.P.
1 mole of CO₂ = 44 g
Therefore;
44 g of CO₂ = 22.4 liters
What about 11 g ?
= (11 g × 22.4 l)÷ 44 g
= 5.6 l
Therefore, 11 g of CO₂ will occupy a volume of 5.6 liters at N.T.P.
