Sebastian V. Mentoring -Medical School Preparation- tutor, Mentoring ...

Sebastian V.

Currently unavailable: for regular students

Degree: Medicine (Bachelors) - Oxford, Oriel College University

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About me

About Me:

I am a 4th year medical student at the University of Oxford. I find the sciences fascinating and enjoy combining knowledge and problem solving in many applications, and hope that I can instil this passion in you.

I have been coaching swimming since I was 15 and during my time at sixth form was a tutor to year 10 and 11 GCSE students, so I have a lot of experience with teaching. At university I have been involved in tutoring small groups of 1st year students, so have a good understanding of communicating higher-level university material.

Tutorials:

I believe that the most important aspect of teaching science is to keep the subject interesting and ensure basic understanding of the topics. During tutorials I will explain concepts using diagrams and analogies, as well as talking through each concept. Most importantly, once we have finished covering each topic you will be capable of explaining it back to me.

You will guide what is covered, how long we spend on each topic, and when we move on.

University Applications:

I have applied to university for studying medicine, and again for clinical school, so understand how difficult it can be. UCAS and personal statements can be very intimidating, and hopefully I can pass on some of my experience to make it easier.

Getting in touch:

If you have any questions please send me an email or book a Meet the Tutor session. If you can, please let me know your exam board and the topics that you hope to focus on.

Thanks! 

Subjects offered

SubjectLevelMy prices
Biology A Level £20 /hr
Chemistry A Level £20 /hr
Biology GCSE £18 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
-Medical School Preparation- Mentoring £20 /hr
-Personal Statements- Mentoring £20 /hr
.BMAT (BioMedical Admissions) Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
ChemistryA-LevelA*
MathsA-LevelA*
PhysicsA-LevelA*
BiologyA-LevelA
BMATUni Admissions Test5.2 Section 1, 7.0 Section 2, 4A Section 3
Medical SciencesBachelors DegreeFirst Class
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

15/01/2014

Currently unavailable: for regular students

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Questions Sebastian has answered

How do I calculate the percentage by mass of a metal within an impure substance?

For this question we will look at Sodium Sulphite (Na2SO3) and its titration against Potassium Permanganate (KMnO4). This requires acidic conditions. In this example we need to know several things: That 1.5g of the impure substance was dissolved and made up to a 100cm3 solution. That 25cm3...

For this question we will look at Sodium Sulphite (Na2SO3) and its titration against Potassium Permanganate (KMnO4). This requires acidic conditions.

In this example we need to know several things:

That 1.5g of the impure substance was dissolved and made up to a 100cm3 solution.

That 25cm3 of this solution was titrated against 35.1cm3 of 0.015

moldm-3 KMnO4-.

Step 1) First we need to balance the equation for the reaction of Na2SO3 with KMnO4.

Use ionic half equations first to work out the change in oxidation state.

Mn04- + 2H+ + 5e- --> Mn2+ + H2O  

In Mn04- manganese has the oxidation state +7 (due to oxygen being -2) and in Mn2+ it has the oxidation state +2. Hydrogen and oxygen do not change oxidation state. This means the oxidation state has changed by -5 overall, so there must be 5e- on the left hand side to balance this.

In the other half equation we see that

SO32- + H2O --> SO4- + 2H+ + 2e-

In SO32- sulphur has the oxidation state +4 and in SO42- has the oxidation state +6. Hydrogen and oxygen again do not change. This means the oxidation state has change by +2 overall, so there must be 2e- on the right hand side to balance this.

From these two equations we can make the full ionic equation, which is written as

2MnO4- + 6H+ + 5SO32- --> 2Mn2+ + 5SO42- + 3H2O

Step 2) Calculate the moles of MnO4- used.

Conc = moles/volume and therefore moles= conc x volume

moles = 0.015 x 0.0371   **be careful to convert units from cm3 to dm3 here**

             = 5.565x10-4

Step 3) Calculate the moles of SO32- reacted.

Using stoichiometric relationships we can see that 2MnO4- reacts with 5SO32-, and therefore

moles SO32- = 5/2 x moles MnO4-

                         = 1.39x10-3

Step 4) Calculate total moles of SO32-

In the information at the beginning of the question it says that the original mass of substance was dissolved in 100cm3, but only 25cm3 was used in the titration. Therefore, to get the total moles we need to multiply by 4.

Total moles = 5.565x10-3

Step 5) Calculate mass of Na present

As there are two atoms of Na per Na2SO3 then the moles of SO32- must be doubled to find the total mass of Na.

Mass = moles x molecular mass

           = 2 x 5.565x10-3 x 23.0

           = 0.25599

Step 6) Calculate the percentage by mass

Percentage = mass of Na/total mass x 100

                       = 0.25599/1.5 x 100

                       = 17.1%

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1 year ago

433 views

How does MRI work?

MRI stands for magnetic resonance imaging and uses the application of a magnetic field and radio frequency photons to manipulate the spin state of protons (hydrogen nuclei). Hydrogen is one of the most abundant atoms in the body as it is a key component of many organic molecules and some inorg...

MRI stands for magnetic resonance imaging and uses the application of a magnetic field and radio frequency photons to manipulate the spin state of protons (hydrogen nuclei). Hydrogen is one of the most abundant atoms in the body as it is a key component of many organic molecules and some inorganic ions.

Protons possess a property called spin, which can be in a particular direction. Inside an MRI machine a very strong magnetic field is applied to the body, and this causes all the protons to align their spin along the axis of the magnetic field. Applying radio frequencies at the same time as the magnetic field causes the protons to change spin and become orientated to a different position, for example a switch from spin up to spin down. When the radio frequency is switched off there is nothing forcing the protons out of alignment with the magnetic field, and as a result the protons will return to their original orientation along the axis of the magnetic field. When the protons return to this spin state they lose energy and this produces photons. Sensors that surround the part of the body being imaged detect these photons, and this detection can be used to generate an image.

It is important to note that photons within different environments behave differently, and this difference is what is used to generate the image.

     H                                                                    H

      |                             O                                      |

H -C – R                    /    \                       H – O – C – R

      |                         H     H                                   |

     H                                                                     H

The bonding partners of hydrogen atoms and the molecules in which they exist cause a difference in the time taken to realign to the axis of the magnetic field, and this produces specific delays for each tissue. The information based upon delays can be used to produce a weighted image in which tissues such as fat, muscle, and bone, and substances such as air and water, can be differentiated between. By using knowledge of anatomy the body structures can be identified and examined.

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1 year ago

417 views

How is skeletal muscle contraction initiated and sustained?

Skeletal muscle contraction consists of two stages. The first is the initiation by the nervous system and the events leading up to myofibril interactions. The second is the actual shortening of sarcomeres, which requires both the presence of Calcium and ATP. The contraction of skeletal musc...

Skeletal muscle contraction consists of two stages. The first is the initiation by the nervous system and the events leading up to myofibril interactions. The second is the actual shortening of sarcomeres, which requires both the presence of Calcium and ATP.

The contraction of skeletal muscle cells is under voluntary control by the nervous system. Motor neurons from the spinal cord terminate at the neuromuscular junction – a structure that acts as a specialised synapse. These nerve cells release the neurotransmitter acetylcholine onto nicotinic receptors. Acetylcholine binding to its receptors causes an efflux of potassium (K+) and influx of sodium (Na+) through an ion channel. These ions cause changes in membrane potential in the same way as action potentials do in nerve cells. However, muscle cells are structured differently and possess specialised T-tubules that descend deep into the cell. T-tubules allow for these changes at the membrane to be passed deeper into the cell, and transmitted to a structure called the sarcoplasmic reticulum (SR). The SR is a storage of Calcium (Ca++), and when stimulated releases this Ca++, which is necessary to allow myofibrils to interact. These steps initiate muscle contraction.

Calcium released from the SR binds to a protein called tropomyosin. This protein normally blocks sites on actin that bind myosin heads during contraction. Adding Ca++ causes a change of shape that reveals these binding sites and allows contraction to begin. Muscle contraction consists of cycling between relaxed state, cocked state, and the power stroke, which depends upon ATP hydrolysis. Provided ATP and Ca++ remain at high enough concentration then the muscle contraction will be sustained. 

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1 year ago

331 views
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