What Is the coefficient in front of Whenllowing half-reacuon balanced acidic solution using the smallest whole number coenclents por FerkIOz (aq) IxkaqhEnter your a...

The following oxidation-reduction reaction occurs in acidic solution. When the equation is balanced using the smallest set of whole-number stoichiometric coefficients possible, what is the stoichiometric coefficient for water? (Hint: In addition to the species shown in the original equation, $\mathrm{H}^{+}(a q)$ and $\mathrm{H}_{2} \mathrm{O}(l)$ can also appear in the balanced eanation ) $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)$$ a. 13 b. 19 c. 11 d. 7 e. 4

So this problem is to balance a read up equation. Um, we have, um the question here that needs to be balanced. So when we want to know, uh, what is the the code revision of the reactant. So s a N B. So what's the ratio? Can we need to balance out to find out? Uh, so to balance a reduction equation, we will need to check the oxidation state. So, uh, so the core in in the the first formula. So Kareen is positive one. And on the right side is negative one and I A dean is zero. And in our date, I also three miners is positive Five. So we'll need to check the increase in the decrease of oxidation state. So ah, for Corey is going from positive one and going down to negative one. So, um, it decreased by two charges. Okay, let's check Aydin. So, um, it goes from zero to positive five. That's for for one. I didn't, Adam, but we have Ah, and I d mark you, which is I two. Which means if we had two ideas atoms, each one increased by five. So we're going to have need to increase by 10. So the increase and the decrease need to be balanced. So we're gonna need to multiply by the negative, too. Uh, bye bye. That way, the number of electrons that transfer are equal. So which means we're gonna need to add a fight here. Ah, for cl all minors. And then we need to add a buy here and then for find in. So we we will have to I Dean Adams. So here we only, uh, need one. So then we don't need to add any co vision. It's one and on right side. Then we will need to at a two. Ah, we can go ahead. Um uh uh, the hydroxide ion and water two balance this equation, or we can stop right here because we already find out. Um, the ratio A to B. So and which is 5 to 1. So the answer is the

This problem has three chemical reactions that we will be balancing in acidic solution. First thing we need to do is identify what the 2/2 reactions are. First half reaction is going to be m N 04 minus going to mn two. Plus it appears that in the book that it was written correctly, there is no such thing as M and two plus, so it should actually be has written m and then two plus Otherwise you're not gonna be able to balance it. No. Well, the main diseases are balanced, so balanced the oxygen's with waters we've got for what? Oxygen on the left hand side will have four water to the right hand side. We just introduced eight hydrogen so at a hydrogen lines to the left hand side, then the balance charges. We need five electrons on the left hand side. So the total sum charges two plus on both sides of the half reaction. Then what is left is the chloride going to the HCL. Oh will balance the oxygen's with water by adding one water to the left hand side. Fellas, the hydrogen by adding one hydrogen ion to the right hand side and then balance the charges with two electrons with total charge on both sides is minus one. Well, some these up and cancel the waters and the hydrogen ions that comment herbal sides. We'll end up with a net 11 hydrogen ions on the left hand side. No, we have are two Emmental for minus and are five seal minuses are five hcl owes are two mn two plus is and then you will see that there are waters. Comment Herbal sides eight on the right hand side five on the left. So we have a net three on the right hand side. This is in our balance street reaction in acidic solution, Then go to the next one. First half reaction is n 03 minus going to end up to bounce the oxygen's with water having an H 20 to the right hand side them Balance the hydrogen is with hydrogen ion spell. It's the charges with an electron, so the total charge on both sides is zero. Then we'll go to the next half reaction, which is what's left eye to going toe. I owe three minus balance the ideas first by making sure that there are two on both sides of the equation. Then we have six oxygen's on the right hand side. So let's six waters to the left hand side. Just introduced 12 hydrogen ions or 12 hydrogen, so add 12 hydrogen ions to the right hand side and then balance the charges with 10 electrons of this, some charge on both sides is zero. Men will multiply the first half reaction by 10 so the electrons will cancel and some them up. We'll end up with a net eight h plus on the left hand side. There's 12 on the right hand side in this one, and then the 10 times to 20. So we have eight, um, net. On the right hand side, we will have our 10 and all three minus are too R two. I owe three minus the 10 ano twos, and we'll end up with just four waters on the right hand side because we have 10. Water's coming from this chemical reaction, and we have six. Water's coming from this chemical reaction. So that's a net four on the right hand side. And for the last one, we've got no tu minus going to you know there's only one reactant in this case. So at O to minus goes to N o and 1/2 reaction and N o to minus goes to an 03 minus in another half reaction for this first half reaction, we need to oxygen's on both sides, so we need to add a water to the right hand side. Then we'll have to hydrogen ions to the left hand side to balance the hydrogen. Glad an electron to the left hand side of the charges or balanced in the total charge on both sides is zero. Then we have our second half reaction I mentioned. And no tu minus goes to, you know, three minus Belus. The oxygen's with water. The hydrogen is with hydrogen ions and the charges with the electrons, we need to multiply the first half reaction by two. So the two electrons will cancel in the second half reaction and then we sum everything up, recognizing that we have a net two hydrogen ions. We've got this times two, which gives us four. Subtract off these two. We just have to on the right hand side. Then we have our three to minus going to to know and are, you know, three. So we're summing up these two right here with this 1st 1 most played by two. That's how we get our three. And then we have our single and no three minus. And we have a net one water on the right hand side, two waters here. One water here. So just one water on the right hand side. This is our balanced Redox reaction. An acidic solution.

Solution for the given problem problem 1 63. The problem is when the equation here given equation is cr 207 minus plus, it's positive Bliss. I negative mm That is see I three positive Bliss. Hi Dean. This balance to give the correct coefficient for I've seen having negative judge. So here it will be cr two or seven to minus plus it's positive and plus um There will be six coefficient for i. d. And this will be this building you chromium three positive plus. Are you Dean To -18- 1812. So We have the correct coefficient is six. Well I. D.

Question Number 72 is a pH calculation of a weak acid. Given the concentration of the weak acid and the K A value of the weak acid. The first thing we need to do is write the chemical reaction for the weak acid reacting with water producing hydro Nia, my on and the conjugate base cielo minus the K expression then will be equal to the hydro knee um, concentration multiplied by the conjugate based concentration, the hypochlorite divided by the concentration of the reactant hippo Cloris acid. Excluding water because it's a liquid because the Stoke eom a tree is one toe one, the concentration of hydro knee um, will be equal to the concentration of the hypochlorite. So okay, which was given to us at 3.5 times 10 to the negative eight will be equal to the hydro knee. Um, concentration squared because thes two concentrations air the same divided by the concentration of the weak acid, which was provided at 0.501 Moeller rearrangement gives us the hydro nian concentration being equal to the square root of 3.5 times 10 to the negative eight multiplied by 80.501 or 1.32 times 10 to the negative four. Moeller Because this is the Hydro Nia Mayan concentration and pH is the negative log of the hydro knee, Um, concentration. We simply have to take the negative log of this value to get a pH of 3.88