For this item, we need to assume that air behaves like that of an ideal gas. Ideal gases follow the ideal gas law which can be written as follow,
PV = nRT
where P is the pressure,
V is the volume,
n is the number of mols,
R is the universal gas constant, and
T is temperature
In this item, we are to determine first the number of moles, n. We derive the equation,
n = PV /RT
Substitute the given values,
n = (1 atm)(5 x 10³ L) / (0.0821 L.atm/mol.K)(0 + 273.15)
n = 223.08 mols
From the given molar mass, we calculate for the mass of air.
m = (223.08 mols)(28.98 g/mol) = 6464.9 g
<em>ANSWER: 6464.9 g</em>
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :
a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
Answer:
1. D (24.0 moles CO2)
2. A (.239 moles H2)
Explanation:
1. First Balance the equation
1 C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
Then set up a stoiciometric equation so that the moles of O2 cancel out
40mol O2 x = 24.0 moles CO2
2. Set up a stoichiometric equation
10 grams Fe x x = 0.239 moles H2
Atoms of the element beryllium would most likely behave similar to the way d lithium behaves