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Lel a € R Causider the system ef mQuaticnsI+20+5: 50+13: SWu + (a? _ 2):0=Find che valuc(s) of a for which thc sysrcm has No slutionsExaetky #u:" slution...

Question

Lel a € R Causider the system ef mQuaticnsI+20+5: 50+13: SWu + (a? _ 2):0=Find che valuc(s) of a for which thc sysrcm has No slutionsExaetky #u:" slution:[nlinite'ky ILAny slutious

Lel a € R Causider the system ef mQuaticns I+20+5: 50+13: SWu + (a? _ 2): 0= Find che valuc(s) of a for which thc sysrcm has No slutions Exaetky #u:" slution: [nlinite'ky ILAny slutious



Answers

The system $$ 2 \mathrm{AgI}+\mathrm{Sn} \rightleftharpoons \mathrm{Sn}^{2+}+2 \mathrm{Ag}+2 \mathrm{I}^{-} $$ has a calculated $E_{\mathrm{ccll}}^{\circ}=-0.015 \mathrm{~V}$. What is the value of $K_{c}$ for this system?

Running this equation for states of the hydrogen atom for which the orbital quantum number L. Is zero is given in your textbook, Verify that equation 39, which describes the ground state of the hydrogen atom is the solution of this equation. Okay, so the proposed way function, sorry, is equal to went over the square root of pi eight of the 3/2. You did? The negative are over A. So let's substitute this into the right side of the schrodinger's equation and we need to show that it's zero. So first we have that the derivative decide er is negative. One over The Spirit of Pi eight of the 5/2, it is the negative are over A. So in the short anger equation we have one over r squared DDR of R squared do you side? Er So it's one over the square root of pi eight of the five half -2 over our Plus one over a eating the negative are over A. Which comparing it to sigh is one over a -2 over our That's one over a say. So the energy of the ground state is negative and either the fourth over 8 ε not squared eight squared. And we know that the border radius is H squared absalon not over high M. E squared. So that gives us that the energy is equal to negative E squared over eight. Hi. Absolutely not A. And the potential energy is negative E squared over four pi. Absolutely not our. So we have eight Pi squared M over age squared E minus you. Time. Cy just eight Hi squared M over age squared E squared over a pie. Absolutely not. I could have won over a plus two over our. Hi. So this is equal to pi M. E squared for age squared. Absolutely not -1 over a plus two over our sigh is equal to one over a -1 over a plus two over our sigh, which we found before. So the two terms are going to cancel, and so the proposed way function to satisfy the schrodinger's equation.

In this problem, just look at it carefully. BB I too, will react in pageants of BSE here, too. ECU was to give compound Are we change which aid 80 cl here, which it A d c L. Here? So this is white PPT. They said white PPT. Therefore, according to the option option, age, correct and said for this problem, A d c l a will be the white PPT in this problem.

In this problem, I can write the DX and ed just look at it carefully at EG tool And or three hole too. Plus to F c l will give actually two Cl two. This is white PPT plus two. HN or three and at EG two CL two in potential two. NS 3 will give at G plus HCG, NH two C L plus energy for CN. And now at EG two CL two in presence of ash to S will form HD plus H D s. This is black in color Plus two SCL. So abs indeed correct here, according to the problem.

Chinese solution when directed with stool and avoided give some right sector own additional access this compound. So could you inform general which hard place here's the correct terms.


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