If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.
We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
Answer:
A meso compound is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
Answer: A
Explanation: The spindle fibers bring the chromosomes together in prophase then apart in telophase.
Answer:
67.5 atm
Explanation:
To answer this problem we can use <em>Boyle's law</em>, which states that at constant temperature the pressure and volume of a gas can be described as:
P₁V₁=P₂V₂
In this case:
P₁ = 101.3 atm
V₁ = 0.500 L
P₂ = ?
V₂ = 0.750 L
We input the data:
101.3 atm * 0.500 L = P₂ * 0.750 L
And solve for P₂:
P₂ = 67.5 atm