Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is
. Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation = ![C_{n} - \frac{\text{monovalent}}{2} + \frac{\text{trivalent}}{2} + 1](https://tex.z-dn.net/?f=C_%7Bn%7D%20-%20%5Cfrac%7B%5Ctext%7Bmonovalent%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Ctext%7Btrivalent%7D%7D%7B2%7D%20%2B%201)
= ![9 - \frac{16}{2} + 1](https://tex.z-dn.net/?f=9%20-%20%5Cfrac%7B16%7D%7B2%7D%20%2B%201)
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.
Answer:
70%
air water and food
carbon dioxide
oxygen
a green pigment
heterotroph
Explanation:
sorry if its wrong i used my best of knowledge i didnt rlly understand most of it
Molarity is expressed as
the number of moles of solute per volume of the solution. The mass of oxalic acid dihydrate needed for the solution is calculated as follows:
Amount in moles: (0.357 mol H2C2O4•2H2O / L) (.250 L ) = 0.0893 mol H2C2O4•2H2O
Amount in mass : 0.0893 mol H2C2O4•2H2O (126.08 g / mol ) = 11.2589 g H2C2O4•2H2O
Hope this answers the question. Have a nice day.
Explanation:
Al2O3
16=molar mass of one oxygen atom
Number of moles: mass/molar mass
Mol of O3: 19÷(16*3) = 0.40 g/mol
mole ratio
0.40: 3 (oxygen)
0.27: 2 (aluminum)
Mass = molar mass*mole
0.27*(27*3)= 21.87g
So,
Given the reaction:
The initial mass of Mg was 0.326, and, in the real experiment, we obtained 0.528 grams of MgO. Right?
Now, let's find the amount of MgO supposed to obtain according to the chemical equation:
Now, the amount that we were supposed to obtain, was 0.54g of MgO. (The result of mutiplying all the previous operations). This is the theoretical yield, what we obtained using the theory.
Now, as you discovered, the mass of MgO in the laboratory was 0.528g so that's the actual yield. This is, what you've found using the experimental process.
Finally, the percent yield can be found using the following equation:
And, we know both values, so let's just replace: