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
Hey there!
The molar mass of magnesium is 24.305 grams.
The molar mass of an element is the same as the atomic mass of an element, but measured in grams.
It's on the periodic table underneath the element symbol.
Hope this helps!
It would be: c/∧ = 3*10^8 / 4.25*10^-11 = 7.05 * 10^18 /s...
<span> The nuclear reactor powers plants to inhale heats that are needed to create/produce steam- (a gas that comes from liquid at its boiling point) The process is called fission. Fission entails the splitting on atoms in the nuclear reactor.
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Answer:
The correct answer is c) 134L
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.
1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K
V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm
V = 133, 86828 l
