Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
For example, the relative rate of a reaction at 20 seconds will be 1/20 or 0.05 s -1, while the average rate of reaction over the first 20 seconds will be the change in mass over that period, divided by the change in time.
Explanation:
Hope it helps u
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Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
Answer:
Via covalent bonds
Explanation:
Atoms in molecules are bonded together via covalent bonds. Covalent bonds are bonds that are formed by sharing of the valence electrons between two atoms.
The atoms can be of the same kind or of different kinds.
- In most molecules, the two atoms are connected by sharing of their valence electrons.
- This way, each atom can attain stability by becoming isoelectric with the nearest noble gas.
- Some molecules are monoatomic, some are polyatomic.
Answer:
10.335
Explanation:
An object was carefully weighed on three different balances
Each of these balances were zeroed before weighing
The masses that were weighed are as follows
10.35 g , 10.355 g, 10.30 g
Therefore the average value of these measurements can be calculated as follows
The total number of mass is 3
= 10.30 + 10.355 + 10.30/3
= 31,005/3
= 10.335
Hence the average value of these measurements is 10.335
