Answer:
The second step would be for formation of solute-solvent intermolecular bonds. Forming bonds always releases energy this would be an exothermic process
Explanation:
Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane =
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield
theoretical yield / 100 %
= 61.0 %
128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
Answer: 0.67 ml
Explanation:
According to the dilution law,
where,
= concentration of stock solution = 226 mg/ml
= volume of stock solution = ?
= concentration of working solution= 15 mg/ml
= volume of working solution= 10 ml
Putting in the values we get:

Thus volume of stock solution needed to dilute to have 10mL of working solution at the above concentration is 0.67 ml