Answer:
Explanation:
Hello.
In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:
Whereas n accounts for the moles which are computed below:
Thus, the entropy turns out:
Best regards.
I believe the answer is 50.5 molecules
For the first one, you have to find one that has both a metal and a nonmetal in it, plus potassium. Bonds between nonmetals are called covalent because they share electrons and a bond between a metal and a nonmetal is an ionic bond because they exchange electrons.
Potassium chloride is KCl. Only an ionic bond.
Potassium hydride is KH. Only an ionic bond.
Potassium nitrate is KNO3. There we go, that has a covalent bond between the nitrogen and oxygen, and an ionic bond between potassium and the nitrogen and oxygen.
For the second one, potassium chloride is the answer because the other three also have covalent bonds. Chloride is the only one that isn't a compound.
Explanation:
when one reactants is in excess, there will always be some left over. The other reactants becomes limiting factor and controls show much of each product is produced.while using excess percentage yields this is at the expense of atom economy.
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL * = 0.295 L