Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Thet N.
Clinical Researcher and Biochemistry Major at Columbia University, New York
Tutor Satisfaction Guarantee
Chemistry
TutorMe
Question:

What is the pH of a 1.00M HF (aq) solution? (Ka=7.2x10E-4)

Thet N.
Answer:

To answer this question, we must first identify that HF is not a strong acid, but rather a weak acid. Therefore, the acid will not dissociate completely, or much at all, which is indicated by the very low value of the Ka constant, also known as the dissociation constant. Next, we write the chemical equation of the process, which involves HF being added to H2O. HF + H2O <---> H3O+ + F- The equation makes H3O+ (We will write this as H+ as a shortcut) because HF is an acid and therefore submits it Hydrogen atom to form F- Next, we must identify the "major species"--i.e. the important species in the equation. Here, the major species are HF and H2O, but because water's dissociation constant (Kw=1x10E-14) is less than that of HF, we recognize that HF produces the bulk of the hydrogen atoms on the right side of the equation. Therefore, we will focus on HF: HF <---> H+ + F- And now we set Ka equal the concentrations of the products over the concentration of the reactants, because this is how the dissociation constant is determined. Ka= 7.2x10E-4 = [H+][F-] / [HF] Because we are dealing with a weak acid, we don't know the concentration of each species at the equilibrium, so we must determine them. Initial: HF=1M, H+=0, F-=0 Change: HF: will lose concentration, because it is a reactant, and the other two will gain. Equilibrium concentrations: HF: 1M-x, H+ = F- = 0+x = x Then, we plug in these variables into the previous equation: Ka= 7.2x10E-4 = (x)(x) / (1-x) here, we solve for x, which is 2.7x10E-2, and plug back into each of the values. Here, x is equal to the concentration of H+, so we have what we need. Lastly, we plug the concentration of H+ into our equation for pH: -log[H+] = pH -log (2.7x10E-2) = 1.57 pH= 1.57 We have found the pH!

Psychology
TutorMe
Question:

How does classical conditioning contribute to the idea of drug tolerance?

Thet N.
Answer:

Classical conditioning suggests that we react to certain stimuli in characteristic manners--for example, that when a dog sees food (unconditioned stimulus), its mouth will automatically begin to water (unconditioned response). Classical conditioning claims that things paired with unconditioned stimuli, called conditioned stimuli (like a light is flashed before a dog is fed), and that the conditioned stimuli will trigger a conditioned response. In this case, showing a light to the dog after conditioning him will make him salivate, even if there is no food present. Classical conditioning and homeostasis both come into play in creating drug tolerance. Homeostasis is essentially our body's tendency to create stable and normal conditions (e.g. blood pressure, body heat, etc.). If a person decides to do heroin, which decreases pain sensitivity and dries up mucous membranes, the body will try to create stable conditions (homeostasis) by increasing sensitivity and increasing mucous after the first time heroin is done. However, this is where classical conditioning comes into play. The body has become conditioned to certain stimuli before doing heroin (like seeing a spoon or needle), so when an addict tries to do heroin again, the environment (conditioned stimuli) will trigger a conditioned response, homeostatic effort (e.g. increased pain sensitivity and mucous). As a result, heroin users' bodies become more and more conditioned to the stimuli and therefore begin to develop psychological drug tolerances. The only way drug tolerance could be avoided is if the stimuli surrounding the user is different every time heroin is done--if the addict went to a different room, stayed with different people, used a different color syringe each time, then there is a possibility that conditioning would not happen, and the body would not develop tolerance; but this is almost never the case. Drug tolerance can create addiction by bringing about cravings when a user sees conditioned stimuli similar to that of the environments that surround their usual heroin routine, be it the time of day, the particular room, or the presence of a particular friend. As a result, the user's body will make a conditioned response, raising pain sensitivity and increasing the activity of mucous membranes, preparing for the use of heroin. But when there is no heroin, the user is left in sometimes excruciating pain and other complications--this is addiction, caused by drug tolerance through homeostatic classical conditioning.

Calculus
TutorMe
Question:

How do I define and find the qualities of an osculating plane? I have trouble picturing it.

Thet N.
Answer:

First, to begin approaching the problem of finding an osculating plane, it is important to know its definition. An osculating plane, in essence, is the closest thing to a plane that defines a particular curve. So, to start, let's find the relevant vectors: the normal, tangent, and binormal vectors from r(t). To find the tangent vector, simply find r'(t) (finding the unit tangent vector is useful, but not exactly necessary when finding the osculating plane--this is a great shortcut to do the problem faster and avoid mistakes!). Once you have found r'(t), you have found T and you can now take the derivative of vector T to find N, the normal vector, also known as r"(t). Now, imagine drawing a curve on your desk. This curve is parallel to the ground. The tangent vector is the same tangent you learned of in earlier levels of calculus, finding the slope at exactly one point. Now, to picture the normal vector (normal to T), imagine drawing a 90 degree angle from the tangent line and drawing that line away from the curve--that is the normal vector N! We can use this to find the binormal vector now. The binormal vector, essentially, is normal to both N and T; imagine a line pointing straight up or straight down at the curve; this is the binormal vector B. To find B quantitatively, we simply take the vectors T and N and take their cross product. Once we find B=TxN, we can start to define the osculating plane. Keep the earlier image in your mind, and understand that the osculating plane is just about parallel to your desk. Now, we use a vector normal to the plane and a point on that plane to define the plane. Because the osculating plane is parallel to your desk, N and T both lie on it; but B does not. B is normal to your osculating plane, so we can use it to define the plane! Now, we finish the problem. Given any point on the osculating plane and having found B, we can plug it into the equation of a plane. Let's say our B vector is <a,b,c> = <1,2,3> and a point (x,y,z) = (4,5,6) is on the plane. The equation of a plane is: a(x-Px) + b(y-Py) + c(z-Pz)=0 (Px, Py, Pz) denote the respective points. So our osculating plane will be: 1(x-4) + 2(y-5) + 3(z-6) = 0.

Send a message explaining your
needs and Thet will reply soon.
Contact Thet
Ready now? Request a lesson.
Start Session
FAQs
What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Session" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.