A student places six plants of equal sizes into six different, equal-sized pots containing the same amounts of potting soil and
labels the pots U, V, W, X, Y, and Z. Using two pots in each of the three experiments, the student varies the amounts of sunlight, water, or air one plant receives but leaves the other pot unchanged. The tables show the factor varied in each pot and the observations made by the student after one week.
Which conclusion best explains the connection among all three experiments?
A. All plants grown in pots require the same amounts of the abiotic factors for their growth and survival.
B. All plants of the same size require the same amounts of the abiotic factors for their growth and survival.
C. The plants require a balance of all abiotic factors for their growth and survival.
D. The plants depend very little on abiotic factors for their growth and survival.
Which conclusion best explains the connection among all three experiments?
A. All plants grown in pots require the same amounts of the abiotic factors for their growth and survival.
B. All plants of the same size require the same amounts of the abiotic factors for their growth and survival.
C. The plants require a balance of all abiotic factors for their growth and survival.
D. The plants depend very little on abiotic factors for their growth and survival.
1 answer:
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N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
(molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams