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$\mathrm{CHCOONa} \stackrel{\text { Electolysis }}{\longrightarrow}[\mathrm{A}] \frac{\mathrm{L}_{2}}{\text { liquid } \mathrm{NH}_{\mathrm{I}}}[\mathrm{B}]$ Here, [B] is (a) $\mathrm{CH} \equiv \mathrm{CH}$ (b) $\mathrm{NaC} \equiv \mathrm{CNa}$ (c) $\mathrm{I}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}$ (d) $\mathrm{IC} \equiv \mathrm{CNa}$
So this grammatical problem is about word choice, and in this case, we can't. So we're gonna cross out. First of all, answer choice, Jay, because they're the specific pronoun it's does not apply to. It doesn't make sense in this context. We're not referring to an object referring to the author narrators referring to herself. So a per the rules of grammar we have to use again. If she's referring to herself, then she has to use the, uh, the phrase that she is using a to moment. We have to keep the answer. We gotta keep the word the same.
In this problem, I'm writing the reactions. So just look at it carefully. OK. FECN six came food, FT CN six plus six. He S CN k s c N will react to form, get to the f e SCN six and now I'm going to write the second reaction. They just look at it carefully Or FECL three. That's pretty careful. F E CN six, then it will give 54 F e CN six plus two old KCL. So option age correct here.
Problem. We're asked to calculate the lap last transform of the function e to the negative t sign to t. And to do this, we're just gonna work directly from the definitions integrating each of the negative x t times using the negative t sign to t t T. Now to integrate this, we're just gonna apply a integration by pots. In this case, we're gonna combine these two terms to be able to e to the one plus s thymus negativity. If we perform the integration by parts anti differentiating this, we're gonna obtain the value negative one of the one plus s corresponding to this negative one plus ask it times E to the negative one plus s T sign to t evaluated at zero. An infinity minus differentiating sign to t gives you to sign coast t. So this equation is e to the negative one plus s t coast to t. Nice thing here is you can also perform an integration by parts to evaluate this side on dhe. This side is completely annihilated because sign of zero zero and each the negative infinity is infinity performing in integration by bus again be obtained. This is equal to negative to over one plus s squared times e to the negative form plus s t coast to t evaluated at zero and infinity plus two times the integral of each of the negative one plus s t coast True T. Now, unfortunately, these terms are annihilated. If we can't fight them will find that the value of infinity is zero. But the value of zero is one to obtain that this is eagle to negative to over one plus s square minus four over one plus s squared times this integral which you're recognize as the black last transform Look e to the negative t coast duty evaluated s which is nice because we've expressed the lack last transform of a function in terms of itself. If you rearrange this equation will get, uh, that's, um, one plus four over one plus s square times the lack life transform with you to the native t sign to t uh, evaluated. It s people to to over one plus x squared. And now dividing on this side and simplifying we obtained the lap. Last transform is gonna be equal to two over one plus X squared plus four
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