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HCCH + CH:CHza.q Solution from a. + D-O-DC. + H `CH:CHzOH0-CH:CHzOH+ H-I <77777777d.HzCCHz +HzSO4 <7777777e....

Question

HCCH + CH:CHza.q Solution from a. + D-O-DC. + H `CH:CHzOH0-CH:CHzOH+ H-I <77777777d.HzCCHz +HzSO4 <7777777e.

HCCH + CH:CHz a. q Solution from a. + D-O-D C. + H ` CH:CHzOH 0- CH:CHzOH + H-I <77777777 d. HzCCHz + HzSO4 <7777777 e.



Answers

For a solution that is $0.164 \mathrm{M} \mathrm{NH}_{3}$ and $0.102 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$ calculate (a) $\left[\mathrm{OH}^{-}\right] ;$ (b) $\left[\mathrm{NH}_{4}^{+}\right] ;$ (c) $\left[\mathrm{Cl}^{-}\right] ;$ (d) $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$.

The equally room in mixture off H N 02 and any, you know, two is represented as follows. We also have the ionization constant off asset. The expression for equilibrium is shown air the initial and equally room constant. The initial and equal room concentration for this system can be summarised into tablet form as represented over here. Substituting in these substituting these value into Kate. That is ionization constant for acid. We have a value for X. That is 1.13 into 10 days. Super minus four. The value off X is lesser than 5% off 0.750 point 0030 hour Assumption over here is valid. The equilibrium, concentration off hide. Roni, Um, I'm in. This solution can be determined from its initial concentration and change in that concentration, as indicated in the last two rule. Over here, off the table, the concentration off hydro near mine turns out to be 1.13 into 10 day super minus four

In this question we have to calculate the concentration of hi johnny on my own in the solution of 0.75 M 18 or two and 0.30 M N A N 02 And the reaction give him and he also given as 4.5 comes into the bar negative five. Now we'll use Henderson. Henderson has luck equation Ph is equal to peek. A means negative lock A Plus law, concentration of a minus. Yet the night dry down divided by a concentration of that. We can sit here. It is net processing. So pts minus law here K. Is given 4.5 time staying to the bar negative file. 4.5 guns tend to be born negative five blast law. 0.30 divided by 0.75 Love minus log 4.5. Into the power minus five using calculator we get 4.35 and for this portion you get minus 0.40 So the ph will be 3.95 Now we have to find out the concentration of fighting a mile. So concentration of hydrogen helium iron is equal to 10 to the power negative ph, which is 3.95 And using the calculator, it comes 1.1 turns tend to the bar negative for so the concentration of the Hibernia man in the given solution is 1.1 times 10 to the power negative for.

That's all for the hydroxide iron concentration in this buffer, a buffer is a mixture of a weak acid or week, or we based on its conjugate salts. So our equation here Stage three on H two with each too old hunt you get here and we'll have to set up and ice table now, knowing that this is point 125 smaller 6.130 Moeller zero minus x x and plus effects 0.1 to 5 minus x Put 130 plus x and x k b we C H three or R K B expression First, which minus or C h three n. H two were given the K B, which is 4.4 times 10 to the negative. Four. Substitute our values in here. We do get a quadratic equation here. So going ahead and solving the quadratic here when I plug this into my equation. Silver 4.4 times 10 to the negative four We have X times 0.130 plus x and in the denominator 0.1 to 5 minus X. Going ahead and solving we get to roots. I'm going to ignore the negative route and I get 4.2 times 10 to the negative four. So X is equal to 4.2 times 10 to the negative four. And if we go back to the ice table, the hydroxide I in malaria T is equal to X there for the O. H minus. Morality for this buffer is 4.2 times 10 to the negative four Mueller.

This question asks you to balance three redox reactions using the half reaction technique. This then requires you to right each reaction As two separate half reactions. For the first one we have F two going to HF The second half reaction is the two plus Going to be three plus. The first thing we do is balance the atoms of everything that's oxygen does not oxygen or hydrogen. So we have to florentines and one floor een. So we need to put a two in front of HF, one Vanadium, 1 Vanadium, So We're Good. Then we balance oxygen's, we don't have any oxygen's. Then we balance hydrogen with hydrogen ion, we have two, hydrogen is on the right hand side, so we need to H plus is on the left hand side. Then we'll balance charges. We have no total charge, no charge two plus. So we need to electrons on the left hand side so that the charges zero on both sides of that first half reaction. Now we have a two plus and a three plus. So to make each side B2 plus by adding electrons will add one electron to the right hand side. So the electrons cancel between the two half reactions when we some them together. Well, multiply that second half reaction by two and the electrons will cancel. And when we sum them up we'll get to H plus goes to F plus F two plus two. The two plus Goes to to the three plus plus two HF. To verify that it's balanced Well balance atoms two hydrogen two hydrogen two floor Eanes to floor Eanes to vanadium seems to vanadium songs. And then what about the charge? No charge? Two times 36 plus. Uh two times 2. four plus. No charge plus two more 6 plus on both sides. So it's balanced for the next one. We have H M n 04 minus Goes to MN two and br minus goes to BR two. So we'll make sure that all the atoms are balanced one Mn, another Mn. So that all atoms except hydrogen and oxygen are balanced will then balance for the top reaction. The oxygen's we have four oxygen's and two oxygen's. So we need to more oxygen's, which we will add as water to the right hand side, that just introduced four hydrogen. So we've got one hydrogen. So we need three more hydrogen is on the left hand side, which because it's an acidic solution will add as H plus, then we'll balance charges, no charges on the right hand side, we've got one minus and three plus, so that's two plus. So we need to electrons so that both sides are not charged. Then for the second half reaction, we've got to bro means and one bro means, so we'll add it to, there's no hydrogen or oxygen so will balance charges by putting two electrons on the right hand side so both sides have a tu minus charge. Then the electrons are equal. So we don't need to multiply either reaction by a constant will simply sum them up and the electrons will go away and this is what we will get To verify that it's balanced. Will recognize we have 34 hydrogen And two times 2 four hydrogen. Then we've got one manganese, 1 manganese, four oxygen's two times one is two plus two more. Four oxygen's to borrow means to bro means. What about the charges? No charge on the right hand side? We've got minus two minus one. That's minus three plus three plus, gives it no charge on the left hand side. And for the last one we've got xenon trioxide going to zen in and selenium to minus going to selenium. All the atoms are balanced except for hydrogen and oxygen. So will balance the oxygen's with water. three, oxygen's no oxygen. So we need 33 waters on the right hand side that introduced six hydrogen. So will introduce six hydrogen ions to the left hand side because it's acidic solution. Then we'll balance charges, no charge to the right hand side, six plus to the left hand side. So we need six minus to make it no charge on both sides. Then we have 2- and no charge. So we need to minus to the right hand side to make sure the electrons council, we're going to have to multiply the second half reaction by three and make sure we distribute it through. When we add everything up, the electrons will cancel. And this is the equation that we'll get to make sure it's balanced. We've got six hydrogen three times to six hydrogen is one scene on one C, none 123 oxygen's three times 13 oxygen's three selenium is three selenium. I don't know what about charge. No charge on the right hand side. Three times to that. Six minus Plus. The six plus gives us no charge on the left hand side.


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