The answer is -3. it would have a negative charge to gain
Answer:
Just make it sound right please it don't even add up what was you even thinking like come on now...lol
Explanation:
What does this mean I don't understand your question.
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:
400 cm³ of ammonia, NH₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.
Finally, we shall determine the maximum volume of ammonia, NH₃ produced from the reaction. This can be obtained as illustrated below:
From the balanced equation above,
3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.
Therefore, 600 cm³ of H₂ will react to produce = (600 × 2)/3 = 400 cm³ of NH₃.
Thus, 400 cm³ of ammonia, NH₃ were obtained from the reaction.