Solubility, fatty acids, and uranium are all topics of the three discrete questions we cover on this next installment of Next Step full-length 10 breakdown!
This is our last set of discrete questions before we move on to our next section. As always, we’re joined by Bryan Schnedeker of Next Step Test Prep.
[01:05] Question 57: Solubility of Bases
The addition of which of the following compounds to water would reduce the solubility of ferrous hydroxide?
- (A) Ammonia
- (B) Ethanol
- (C) Acetic acid
- (D) Trichloroethane
There are three scientific principles to remember.
- Recognize that hydroxide is a base (OH-). Anything that creates hydroxide is a base. And ferrous hydroxide is very sparingly soluble. It’s not much of a base but technically it is.
- Bases would interfere with each other trying to dissolve, so they would reduce each other’s solubility. Next fact to remember is that for solubility rules, acids and bases like to dissolve each other. They help each other dissolve. Since the OH and the H get together to make water. In that same vein, bases interfere with each other. So if you try to dissolve multiple bases, they’re going to make each other harder to dissolve. Acids interfere with each other due to the common ion effect.
- Step 3 is just to know which of the answer choices is a base. Ammonia is a weak base. When the MCAT wants to give a weak base, 8 times out of 10, they use Ammonia if they want a weak base so this is the right answer here.
[03:35] Question 58: Cell Membrane Forces
Both eukaryotes and prokaryotes possess desaturases which can increase membrane fluidity by adding a cis double bond to a fatty acid chain on a phospholipid. What’s the primary intermolecular force acting between adjacent fatty acid chains in a cell membrane?
- (A) London dispersion
- (B) Hydrogen bonding
- (C) Dipole-dipole
- (D) Covalent bonds
It says intermolecular force so you can eliminate covalent bonding since it’s an intramolecular force. Fatty acid chains are just big long chains of hydrocarbons. It means the only intermolecular force that predominates for them is London dispersion.
Hydrogen bonding requires hydrogen bonds to be bonded to fluorine, oxygen, or nitrogen. None of which is true in the tail part of the fatty acid chain. And dipole-dipole interactions require actual dipoles.
[05:44] Question 59: Percentages and Conversion
Naturally mined Uranium consists of two isotopes U-235 and U-238. U-238 is more stable than U-235 and cannot be used for nuclear energy. An industrial enrichment process increases the abundance of U-235 from 0.9% to 3.7% in a 500-gram sample of Uranium. How many additional moles of U-235 are yielded as a result?
- (A) 7.8 x 10-2 moles
- (B) 5.9 x 10-2 moles
- (C) 7.8 x 10-1 moles
- (D) 5.9 x 10-1 moles
So there are two steps here. Take that percentage increase on the 500-gram sample and then divide the grams by molecular weight to get moles.
Doing some math, 500g times 2.8% (3.7-0.9) is 14 grams and converting this to moles (14/235) and the answer is B.
You can also do rounding off just to make the math easier. But you also want to round something where you can reduce the fraction. Then try to round so that the denominator of the fraction gets as close as you can to either 100 or 50 or 25. Something you can just scale to 100 to make it a percent. This is a very common technique for manipulating fractions without a calculator on test day.
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