Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
1. Which law is associated with inertia?
2. If you increase the force in an object what happens to the acceleration?
3. If you use the same force on a less massive object what happens to the acceleration?
4. Which law states force is dependent on the mass and acceleration of an object?
5. What causes an object to slowdown or speed-up?
HOPE IT HELPS YOU
Answer:
Explanation:
522 g
Explanation:
Your starting point here will be the balanced chemical equation for this combustion reaction
4
P
(s]
+
5
O
2(g]
→
2
P
2
O
5(s]
Notice that you have a
4
:
5
mole ratio between phosphorus and oxygen. This means that, regardless of how many moles of phosphorus you have, the reaction will always need
5
4
time more moles of oxygen gas.
Use phosphorus' molar mass to determine how many moles you have in that
93.0-g
sample
93.0
g
⋅
1mole P
30.974
g
=
3.0025 moles P
Use the aforementioned mole ratio to determine how many moles of oxygen you would need for many moles of phosphorus to completely take part in the reaction
3.0025
moles P
⋅
5
moles O
2
4
moles P
=
3.753 moles O
2
A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements. If only one atom of a specific type is present, no subscript is used. For atoms that have two or more of a specific type of atom present, a subscript is written after the symbol for that atom.
