2.168 L of air will leave the container as it warms
<h3>Further explanation</h3>
Given
V₁=2.05 L
T₁ = 5 + 273 = 278 K
T₂ = 21 + 273 = 294 K
Required
Volume of air
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature
Input the value :
V₂=(V₁.T₂)/T₁
V₂=(2.05 x 294)/278
V₂=2.168 L
Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M
calculation
molarity =moles of solute/ Kg of the solvent
mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol
moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles
calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol
mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg
molarity is therefore = 0.06/0.09842 = 0.6 M
