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:
whats the question??? I am familiar with brainpop but if u tell me the question ill give u the asnwer
Explanation:
The E°half-cell of the (3) is 0.33 volts .
Given,
(1) Fe3+(aq) +e =Fe2+(aq)
(2) Fe2+(aq) +2e = Fe(s)
(3) Fe3+(aq) +3e = Fe(s)
We know ,
E° half-cell of (1) is 0.77 volts .
E° half cell of ( 2) is -0.44 volts .
By resolving or adding equation 1 and 2 we will get (3) ,
Thus , the value of the E° half cell of (3) is given by ,
E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts
Hence the E° half-cell of ( 3 ) is 0.33 volts .
<h3>What is cell potential? </h3>
It is the difference between the electrode potentials of two half cells.
it also defined as the force which causes the flow of electrons from one electrode to the another and thus results in the flow of current.
Ecell = E( cathode) - E ( anode)
<h3>What is electrode potential ? </h3>
The tendency of the electrode to lose of gain electrons is called as electrode potential.
Learn more about cell potential here :
brainly.com/question/19036092
#SPJ4
Answer:
d, 40 dm3.
Explanation:
According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).
From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.
Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).
With that being said, all H2 has reacted, meaning there's no volume of H2 left. 2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)
1 mole of N2 has reacted, meaning from the 30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.
So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.
D. Not all mixtures are heterogeneous
