The substance that gets dissolved is the solute and the substance that dissolves the solute is the solvent. Example: water is a solvent and salt is a solute
As can be seen in the attached image, α-pyrone has a highly electrophilic carbon atom, since it is attached to two oxygen atoms that are electronegative and subtract electrical charge from the carbon, leaving it with a <u>positive partial charge</u>. By virtue of the above, <u>the bromine atoms, which have an important electron density that makes them good nucleophiles, will be attracted to the aforementioned carbon due to their positive charge</u>, thus favoring the substitution product to a greater extent than that of addition.
Answer:
2NaF + 1Br2 -> 2NaBr + 1F2
Explanation:
2NaF + 1Br2 -> 2NaBr + 1F2
I simply looked at both the left side and right side and saw there was only 1 F on the left side while there were 2 on the right side.
This means we have to make the left side have 2 F by multiplying NaF by 2.
However, this means there are now 2 Na as well. Going back to the right side, I multiplied NaBr by 2 so that there are 2 Na on both sides.
Now there are 2 Br on the right side. Checking back to the left side, there are also 2 Br, meaning this equation is now balanced.
Answer:
1. It is stoichiometric.
2. O2 is the limiting reactant.
3. 9.0 g of C2H6 remain unreacted.
4. 17.6 g of CO2.
5. 85.2%.
Explanation:
Hello there!
In this case, for the given chemical reaction:
We can see that:
1. It is stoichiometric and is balanced because the reactants yields the products according to the law of conservation of mass.
2. In this part, it is possible to calculate the moles of ethane by using its molar mass:
And the moles of oxygen by knowing that one mole is contained in 22.4 L at STP:
Thus, by calculating the moles of carbon dioxide product by each reactant, we can identify the limiting one:
Thus, since oxygen yields the fewest moles of CO2 product, we infer it is the limiting reactant.
3. In this part, we calculate the mass of C2H6 that actually react first:
Thus, the leftover of ethane (C2H6) as the excess reactant is:
4. Since 0.4 moles of carbon dioxide were produced, we use its molar mass to calculate the mass as its theoretical yield:
5. Finally, the percent yield is gotten by dividing the actual yield by the theoretical one:
Best regards!
Its D for the second energy level, but the first energy level is 2.