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Tutor profile: Saurabh S.

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Saurabh S.
i am co-founder and senior UI/UX developer at Flying Homingos
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Questions

Subject:Chemistry

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Question:

Calculate the emf of the following cell: Mg(s)|Mg2+(aq)||2H+(aq)|H2(g)

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Saurabh S.

In order to calculate the emf of the above cell, we first should determine the substance undergoing oxidation and the substance undergoing reduction, and then we write the equations for oxidation half cell and reduction half cell separately. Now, in order to remember, I suggest a mnemonic "LEOGER", which stands for : Loss of electrons = Oxidation, Gain of electrons = Reduction, i hope this helps removing your confusion and helps you remember this. So now as we can see in the equation of cell, Mg (Magnesium) goes from a solid phase into an aqueous phase by attaining a two unit positive charge, in other words, by losing 2 electrons. Hence, Mg undergoes Oxidation. Whereas, 2 H+ ions gain two electrons (one electron each) and change from an aqueous state to a gaseous state forming one molecule of Hydrogen (H2). Now we write the half equations for cell as follows: Mg (s) → Mg2+ (aq) + 2 e- [equation for oxidation always reflects electron release that means loss and is thus, shown on right hand side] 2 H+ (aq) + 2 e- → H2 (g) [equation for reduction always reflects electron gain and is thus, shown on left hand side] In order to find out the emf for the cell, we need to know the standard reduction potential for the half reactions. For reduction half reaction : E0 = 0.0000 V since hydrogen is taken as a reference while calculating standard reduction potential of other electrodes using the SHE (standard hydrogen electrode) For oxidation half reaction, in order to get standard reduction potential, we simply revert the equation and find out the E0 for it. Thus, Mg2+ (aq) + 2 e- → Mg (s) This reaction has a E0 = -2.372 V. E0 (oxidation) = - E0 (reduction) = - (-2.372) V = +2.372 V Now, E0 (cell) = E0 (reduction of 2H) + E0 (oxidation of Mg) = 0.000 V + 2.372 V = +2.372 V Having a positive emf implies that the cell must be a galvanic cell and thus, this electrochemical cell will derive energy from the electrolytic reactions taking place inside it.

Subject:Algebra

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Question:

If (2 + x)^5 + (3 + 2x)^5 = ax^5 + bx^4 + cx^3 + dx^2 + ex + f Find the values of a, b, c, d, e and f.

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Saurabh S.

The above is a question that easily be solved using our knowledge of Binomial theorem. Now, binomial theorem simply states that (x+y)^{n}={nC0}x^{n}y^{0}+{nC1}x^{n-1}y^{1}+{nC2}x^{n-2}y^{2}+....so on upto..+{nC(n-1)}x^{1}y^{n-1}+{nCn}x^{0}y^{n}, Now for the above question we can solve for (2 + x)^5 and (3 + 2x)^5 separately and then add the coefficients of similar power of x and compare them termwise in order to get the values of a, b, c, d, e and f. First, we solve for (2 + x)^5 According to the theorem, (2 + x)^{5}={5C0}2^{5}x^{0}+{5C1}2^{4}x^{1}+{5C2}2^{3}x^{2}+{5C3}2^{2}x^{3}+{5C4}2^{1}x^{4}+{5C5}2^{0}x^{5} Now in order to find the values of 5C0, 5C1, 5C2, 5C3, 5C4 and 5C5, we can either simply use a calculator or take help from a pascal's triangle as follows: 0Cr 1 1Cr 1 1 2Cr 1 2 1 3Cr 1 3 3 1 4Cr 1 4 6 4 1 5Cr 1 5 10 10 5 1 where r is from 0 to n value. Thus, using pascal's triangle, 5C0 = 1 (first element of 5Cr row), 5C1= 5 (second element of 5Cr row), 5C2= 10 (third element of 5Cr row), 5C3= 10 (fourth element of 5Cr row), 5C4= 5 (fifth element of 5Cr row) and 5C5= 1 (last element of 5Cr row) Substituting these values in above equation: (2 + x)^{5}=1*2^{5}x^{0}+5*2^{4}x^{1}+10*2^{3}x^{2}+10*2^{2}x^{3}+5*2^{1}x^{4}+1*2^{0}x^{5} Thus, (2 + x)^{5}= 32 + 80x + 80x^{2} + 40x^{3} + 10x^{4} + x^{5} Now we solve for (3 + 2x)^5: (3 + 2x)^{5}={5C0}3^{5}(2x)^{0}+{5C1}3^{4}(2x)^{1}+{5C2}3^{3}(2x)^{2}+{5C3}3^{2}(2x)^{3}+{5C4}3^{1}(2x)^{4}+{5C5}3^{0}(2x)^{5} Now, in the similar manner as above, we substitute the values of 5C0, 5C1, 5C2, 5C3, 5C4 and 5C5 and solve further: (3 + 2x)^{5}= 1*3^{5}(2x)^{0} + 5*3^{4}(2x)^{1} + 10*3^{3}(2x)^{2} + 10*3^{2}(2x)^{3} + 5*3^{1}(2x)^{4} + 1*3^{0}(2x)^{5} Thus, (3 + 2x)^[5]= 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5 Now, we add the values of (2 + x)^5 and (3 + 2x)^5 ( adding coefficients of terms with same exponent of x) : (2 + x)^5 + (3 + 2x)^5 = 32 + 80x + 80x^{2} + 40x^{3} + 10x^{4} + x^{5} + 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5 (2 + x)^5 + (3 + 2x)^5 = 275 + 890x + 1160x^2 + 760x^3 + 250x^4 + 33x^5 Thus, a = 33, b = 250, c = 760, d = 1160, e = 890, f = 275.

Subject:Physics

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Question:

Explain the Young's Double Slit experiment and his observations.

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Saurabh S.

In the experiment an opaque screen and a monochromatic light source were kept at a distance D from each other. Now the light source was covered with an opaque sheet with two slits (for simplicity, let us assume that the slits were made vertically, although the results will be the same for horizontal slits too) at a distance of d from each other. Let the mid point between the slits be assumed as O and the point corresponding to it on the screen be X. Let the points corresponding to the two slits on the screen be E and F, also slits can be regarded as Slit S1 (the upper one) and S2 (the lower one). Young in his experiment observed that the waves from the two slits produced interference patterns which superimposed on each other and produced alternating dark and bright bands on the screen. This experiment helped conclude and highlight the wave nature of light as it pointed out to interference of waves (light) taking place. His observations of bright and dark patterns can be explained as follows. Now considering S1S2 = EF = d, and S1E=S2F=D. Let us consider a point P on the screen which is at a distance y from the center of the screen X. (Let's assume that this point is above E, results remain the same for any other position too so one need not worry) Now consider triandle S2PF and let us apply pythagoras theorem that is (Hypotenuse) ^2 = (Base)^2 + (Altitude)^2 Thus, (S2P)^2 = (S2F)^2 + (FP)^2 Now we will substitute the values as mentioned above. (S2P)^2 = [D^2 + (y + d/2)^2] ( Do remember that d/2 is distance of F from X and y is the distance of P from X, thus, total distance FP = y + d/2) S2P = [D^2 + (y + d/2)^2]^0.5 = D [ 1 + {(y + d/2)^2 }/D^2 ]^0.5 [We just took a Multiplied and Divided by D so don't worry, the expression does not change, and it will get simpler as you go through. :) ] Applying the pythagoras theorem on triangle S1PE, keeping in mind that PE = y - d/2 as distance EX= d/2 and PX is y. (S1P)^2 = (S1E)^2 + (EP)^2 Doing similar rearrangement as above: (S1P)^2 = [D^2 + (y - d/2)^2] S1P = [D^2 + (y - d/2)^2]^0.5 = D [ 1 + {(y - d/2)^2 }/D^2 ]^0.5 Now let us subtract pathlengths of wave 1 and wave 2 that is subtracting S1P from S2P: D [ 1 + {(y + d/2)^2 }/D^2 ]^0.5 - D [ 1 + {(y - d/2)^2 }/D^2 ]^0.5 Now we know that distance between the screen and slits, D is relatively very much greater than the distance between slits d, thus, (y + d/2)^2/D^2 and (y - d/2)^2/D^2 is very much lesser than 1 as the denominator is much larger than the numerator. Hence, we can apply our simplicity friend, Binomial theorem [ (1 + x)^n = 1 + nx for x<<<1 ] S2P - S1P = D [ 1 + 0.5{(y + d/2)^2 }/D^2 ] - D [ 1 + 0.5{(y - d/2)^2 }/D^2 ] Subtracting and taking D^2 out of the bracket as D, we get: S2P - S1P = 1/2D [ 4 yd/2] = yd/D In order to get bright fringes which are formed because of constructive interference of the waves from the two sources, pathdifference S2P - S1P should be an integral multiple of wavelength \$\$lambda\$\$ Thus, S2P - S1P = n* \$\$lambda\$\$ where n can be any integer ( Remember that negative integers cover for the points below E and hence, this formula is applicable as it is for the entire screen ) Hence, yd/D = n\$\$lambda\$\$ y = n*D*\$\$lambda\$\$/d For all the points at a distance y as above from the center, bright fringes will be formed. Now, if you notice the locus of the points will be a circle as the distance from center of the screen X is constant. Thus, bright and dark alternate circular fringes about the center were observed as concentric circles in the experiment.

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