The molecular formula of the unknown compound is determined as C₉H₁₁NO₂.
<h3>Molecular formula of the compound</h3>
The molecular formula is calculated as follows;
CHNO + O₂ ------------> CO₂ + H₂O
Mass of carbon, C: = (6.64 x 12)/44 = 1.81 g in 2.77 g sample
Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample
Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g
Mass of oxygen, O: = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g
<h3>molar ratio of the elements: </h3>
C = 1.81 g = 0.15 mol
H = 0.186 g = 0.186 mol
N = 0.234 g = 0.017 mol
O = 0.54 g = 0.0337 mol
divide through with the smallest number of moles (0.017 mol);
C = 9
H = 11
N = 1
O = 2
Molecular formula = C₉H₁₁NO₂
Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol
Thus, the molecular formula of the unknown compound is determined as C₉H₁₁NO₂.
Learn more about molecular mass here: brainly.com/question/21334167
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Answer:
Bond energy of carbon-fluorine bond is 485 kJ/mol
Explanation:
Enthalpy change for a reaction, is given as:
![\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%28E_%7Bbond%7D%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%28E_%7Bbond%7D%29_%7Bj%7D%5D)
Where 
and 
represents average bond energy in breaking "i" th bond and forming "j" th bond respectively.
and 
are number of moles of bond break and form respectively.
In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed
So, 
or, 
or, 
So bond energy of carbon-fluorine bond is 485 kJ/mol
Answer:
The purpose of all seed is reproduction and is to grow a new plant, thus propagating the species.
Answer:
390 m
Explanation:
Martha ran 420 m
Cole ran 60 m less than Martha = 420 m - 60 m = 360 m
The distance Kate ran is right in the middle of Martha and Cole's distances
Kate distance = (420 + 360) / 2 = 780 / 2 = 390 m
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
