The grams of N2 that are required to produce 100.0 l of NH3 at STP
At stp 1moles = 22.4 l. what about 100.0 L of NH3
= 100 / 22.4 lx1 moles = 4.46 moles of NH3
write the reacting equation
N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2
mass = moles x molar mass
= 2.23moles x 28 g/mol = 62.4 grams
<span>Glycerol is a sub-unit molecule of a Lipid
So, option C is your answer.
Hope this helps!
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False. elements in the same period have the same number of shells while elements in the same group have the same number of valence electrons.
1 is seconds and meters m/s
2 is seconds and meters m/s^2
3 Newton kg/m
4 Kilograms
Hope this helps!
Answer: 83%
Explanation:
The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.
