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Jacob D.

Biomedical Engineering Student at UT Austin

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Organic Chemistry

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Question:

Which of the following is the STRONGEST acid? 1) CH3COOH 2) BrCH2COOH 3) FCH2COOH 4) ClCH2COOH

Jacob D.

Answer:

The correct answer is 3. This question tests your knowledge of the Inductive Effect. This effect has to deal with both the electronegativity and proximity of the electronegative atom to the acidic group of the molecule. The closer the electronegative atom is to the acidic functional group, the stronger the acid is. The stronger the electronegativity of the atom affecting the acidic functional group, the stronger the acid is. This is because the electronegativity of the atom pulls electrons away from the negatively charged oxygen when the acid is in it's aqueous, ionic form. This creates stability in the molecule, and a more stable acid is a stronger acid. This is also why the proximity of the electronegative atom is important, because the closer it is to the ionized oxygen, the easier the atom can pull electrons from it and stabilize the molecule. We notice in the above question that the first answer choice is wrong, because it does not have an electronegative atom like the other 3 options do. So we can conclude it's the weakest of the 4. The last three all have halogens, a highly electronegative functional group, in the same position and proximity to the carboxylic acid functional group (-COOH). So now that we have determined that their proximity is the same, the answer boils down to which of the 3 halogens has the strongest electronegativity. The answer to this question is Fluorine, so the correct answer is 3.

Biology

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Question:

What is the driving force behind passive membrane transport? Does this type of transport require any input of ATP?

Jacob D.

Answer:

A concentration gradient. No, because the difference in the concentration gradient will cause molecules to undergo diffusion across the membrane without the need for a catalyst like ATP.

Algebra

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Question:

How many solutions does the below system of equations have? 5x − 6y= −25 4x − 3y + 20 = 0

Jacob D.

Answer:

I like to start these types of problems by putting all constants on the same side of the equation. Thus, we have: (1) 5x - 6y = -25 (2) 4x - 3y = -20 I do this because it gives us a general frame of reference about the value of each variable. We can see that removing one x and adding 3 y from the first equation, which yields the second equation, only adds a value of 5 to the constant on the other side. Thus we can conclude that: (3) -x + 3y = 5 There are many ways to solve systems of equations, and in each situation one may be faster than the other. I personally prefer to use substitution, because this is the method that works consistently and candidly in higher level mathematics, so it's a good tool to practice with early on. Substitution involves solving for one variable of the system in any of the provided equations, and then plugging in that solution to another equation in the system. This might seem like a tongue-twister viewing in words alone, so let's do it: First, we will solve for x in terms of y, using equation 2. (Hint: You may also use equation 3!) 2) 4x - 3y = -20 4x = 3y - 20 (we added 3y to both sides) x = (3y - 20)/(4) (we divided both sides by 4) Simplify: (4) x = (3/4)y - 5 Now that we know what x is in terms of y, we can simply substitute it into any of the above equations, and solve for the value of y. Substitute x for (4) into equation (1). 5x - 6y = -25 5( (3/4)y - 5) - 6y = -25 (15/4)y - 25 - 6y = -25 (15/4)y - 6y = 0 (We added 25 to both sides) (15/4)y - (24/4)y = 0 (we converted 6 to fraction form, with denominator 4) (-9/4)y = 0 y = 0 (divide both sides by (-9/4)) Now that we know y is equal to zero, we can plug that value for y back into equation (4). (4) x = (3/4)y - 5 x = (3/4)(0) - 5 x = -5 So, given this system, we see that our math asserts x = -5, y = 0. Let's check our work. (1) 4x - 3y = -20. Plugging in our values for x and y should give us -20. 4(-5) - 3(0) = -20 -20 = -20 Great! Because the question asks how many solutions there are for this system, and we only found one solution that works, we can determine that this system has only one solution combination of x and y that will satisfy both equations 1 and 2.

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