# Blueprint MCAT Full-Length 1: Discrete 2 — Chem/Phys II ## Session 189

We’re talking about gravity, diatomic gases, bond angles, and circuits in discrete 2 in the Blueprint MCAT full-length 1. Follow along and test your knowledge!

We’re joined by Phil from Blueprint MCAT, formerly Next Step Test Prep. If you would like to follow along on YouTube, go to premed.tv.

Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.

## [03:44] Question 27

A book rests horizontally on a table. The book experiences a gravitational force of mg due to the earth’s gravity. According to Newton’s third law:

(A) the book experiences a normal force of mg pushing up due to the table.

(B) the earth experiences a gravitational force of mg from the book.

(C) the table exerts a gravitational force of mg on the earth.

(D) the table exerts a gravitational force of mg on the earth.

Thought process:

This is one of the questions that is probably one of the most missed questions on this exam. Actually, more than one of these answers is a true statement.

But the question is very specifically asking according to Newton’s third law, which is that every action has an equal reaction.

It’s like e=mc2. Everyone knows that equation, but very little people know that’s for calculating mass defect of combining nuclei.

An important thing with reactionary forces, which is what Newton’s third law is about is that they have to be between the same two objects, and it’s got to be the same two types of forces.

So if charge A pulls on charge B with electrostatic traction, then charge B pulls on charge A with electrostatic attraction.

It’s got to be the same type of forces and it’s got to include the same two objects.

Looking at what’s going on here in the question, they say the book is experiencing gravitational force due to the earth. So the earth is pulling the book down, which means with gravity. So it means the book must be pulling up on the earth with gravity. Hence, the correct answer here is B.

A is right. The table is pushing up on the book, and that’s what’s stopping the book from rocketing through the table or speeding up into the sky. So the normal force and Earth’s gravity force are equal and opposite, but they’re not reactionary forces. Because reactionary forces have to be including the same objects and the same types of forces.

A lot of students end up just looking at this like it’s true. And then they pick A. 60% of students pick A for this one because it’s just a true statement and they just read through that. But we’re looking specifically for a reactionary force.

## [12:05] The Difference Between Normal Force and Gravitational Force

The normal force is how hard the table is pushing up on the book. Imagine that I was laying on the floor and you stood on my stomach. Gravity is pulling you down. And I am pushing you up. So the normal force would be me lifting you off the ground and applying that upward force. As much as I’m lifting you up, you’re smashing down on me. And that’s a contact force.

Now, the Earth is pulling you down, and you’re also pulling the earth up, which seems weird, but the earth is huge. Just because the earth is so massive, pulling on it a little bit it’s not going to make much of an effect on its movement at all.

Question 27 seems like a tricky question but it’s a very specific definition question. Therefore, you have to know Newton’s third law. You have to know equal-opposite, but also equal-opposite of the objects and of the type of force.

So reactionary force is if one’s a gravitational force, the other one’s a gravitational force. If one’s electrostatic, the other one’s electrostatic. If one’s contact, the other one’s contact.

## [14:08] Question 28

Nitrogen primarily exists in the atmosphere as a diatomic gas. Which of the following is true about this form of nitrogen?

(A) The presence of a lone pair of electrons on each nitrogen atom in the molecule allows it to act as a strong Lewis base.

(B) The triple bond of electrons creates a region of high electron density that allows N2 to be very reactive as a nucleophile.

(C) Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions.

(D) Atmospheric nitrogen reacts spontaneously with carbon dioxide, which keeps atmospheric CO2  levels at a relatively low 0.04% (on a molar basis) of the atmosphere.

Thought process:

When we breathe in, 70% of the air we breathe is actually nitrogen gas. Oxygen is 20% and that’s the next highest. After that is CO2, then everything else is the last 5%. Most of the air we breathe is nitrogen. It doesn’t make any effect at all.

Diatomic nitrogen is going to be pretty inert because the nitrogens are held by a triple bond. And so B is correct that there’s a triple bond between these two nitrogen atoms. But that makes it not reactive because it’s just really tightly held together. Those things don’t want to interact with other things, which helps you eliminate B and D, because it’s really unreactive.

Nitrogen is not going to be a strong base. If that was the case, then we would breathe in nitrogen gas and our body would get chemical burns from the alkaline activity of the air. That wouldn’t be comfortable.

A, B, and D really highly reactive nature. And that’s not going to be the case. We don’t even notice nitrogen in the air.

This hurts a lot of students as they’re prepping for the MCAT because they’re used to an undergrad reading. If they’re tested on something, they’re expected to just know it. Whereas the MCAT is a lot of puzzling.

## [18:02] Question 29

Which of the following is closest to the bond angle between the carbons in a molecule of acetone?

(A) 90°

(B) 109.5°

(C) 120°

(D) 180°

Thought process:

Breaking down the word acetone, acet- refers to acetyl group. It means two carbons with a carbonyl attached to it. And so that’s where we have acidic acid, which is vinegar, acetyl coA, which is a big part of metabolism.

As you get into the like molecular genetics, we’ll talk about like acetylating histone, which just means adding an acetyl group to it. Then acetone, obviously, as well. And so all of these have the same basis in terms of molecules.

Knowing what an acetal group looks is useful because it shows up all over the place. But just as important as the second half of that word, which is -one, which just means ketone.

So if you have a ketone, that means you’ve got a double bond oxygen in the middle of a compound. And if I’ve got acetone, that tells me that I’m going to have a carbon double bonded to an oxygen.

Now, carbon normally has four bonds. And so if one of those bonds is a double bond, then it’s going to have two other bonds. This acetone molecule is just going make it to three things total, the oxygen and then two other carbons. After that, you can just use geometry to figure it out.

It’s probably going 360 degrees like a peace sign as this orients. So it’s going to be about 120 degrees between each one. So even if you’re unsure of the structure of acetone, and let’s say you didn’t even know the acetyl group thing, you definitely want to know the second part of the word, -one.

That happens big with ketone bodies and metabolism. Any functional groups involving carbons, oxygens, and nitrogens, you want to be really comfortable with carboxylic acids. If you see anything that ends in -oic acid, like propionic acid or acetic acid.

You want to know those structures just because carbon, oxygen, and nitrogen are what we build our bodies out. The MCAT cares a lot more about those functional groups.

## [21:18] Question 30

A circuit is constructed with 12-V battery and four identical resistors, each with a resistance of 16 Ω, hooked up in parallel. What is the total power dissipated by the circuit?

(A) 4 W

(B) 4 J

(C) 18 J

(D) 36 W

Thought process:

If you know the difference between a watt and a joule, you’re down to a 50/50. Worst case scenario, which is a great place to be if you just know your units.

Power is measured in watts. That means it’s got to be A or D. So. Power is a measure of energy change over time. So if I have a 60-watt light bulb, that means it’s burning 60 joules every second.

So what is just a joule per second? So B and C are both joules that’s not power. Power is a joule per second. And so you can get LED bulbs, which are 17 watts and so they burn about one-fourth the energy of your normal 60-watt light bulb because they run 17 joules every second. And so just off of that, we think the answer is probably A or D.

Joule is a measure of energy. So however many energy, the photons are being released. There’s some amount of energy there that’s being put out. Especially if it’s not an LED bulb, then there’s heat released as well as energy. And that’s why the light bulbs get hot. The power is a joule per second, so it’s like miles per hour. That’s definitely a difference between those.

And so the other half of this is the circuitry stuff. So even if we know we’re looking for watts, we’re down to A or D. At this point, we got to figure out the total power.

Power = IV, which is current times voltage.

We’ve got to figure out what the current is, the total current of the circuit. They’ve given the voltage. That makes it easier. So we just need to find the current and multiply that by the voltage.

The issue is we don’t have the total resistance of the circuit. If we had the total resistance, we could just use V = IR, and we’d be done. But we have four resistors. And so we got to figure out what the total resistance of the circuit is.

Because all of this is in parallel, your equation for resistance is 1/R. The total resistance is equal to 1/R1 + 1/R2 + 1/R3 + 1/R4. So the equation is going to basically break down to 1/R = 1/16 + 1/16 + 1/16 + 1/16. So our actual resistance 1/R is equal to 1/4. So that means our resistance has to be 4. And so our total resistance for this circuit is going to be 4 ohms.

And then you can use P = IV going through this and we have V = IR to find the current. So the V=12, the R=4, that means the current is going to be 3. P=IV so that’s going to be 3×12.

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