Answer:
Twice as much.
Explanation:
That's because the freezing point depression depends on the total number of solute particles.
C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)
0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).
That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.
Answer:
Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.
Explanation:
yan na answer
Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>
In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.
See figure 1
I hope it helps!
answer: D
Here is a list of the most common ways to speed up a chemical reaction
Increase the temperature (reactions that absorb energy)
Decrease the temperature (Reactions that release energy)