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Balance the chemical equations in Exercises $5-10$ using the vector equation approach discussed in this section.[M] The chemical reaction below can be used in some industrial processes, such as the production of arsene (AsH ). Use exact arithmetic or rational format for calculations to balance this equation.MnS $+\mathrm{As}_{2} \mathrm{Cr}_{10} \mathrm{O}_{35}+\mathrm{H}_{2} \mathrm{SO}_{4}$$\quad \rightarrow \mathrm{HMnO}_{4}+\mathrm{AsH}_{3}+\mathrm{CrS}_{3} \mathrm{O}_{12}+\mathrm{H}_{2} \mathrm{O}$
$16 \mathrm{MnS}+13 \mathrm{As}_{2} \mathrm{Cr}_{1} 0 \mathrm{O}_{3} \mathrm{5}+374 \mathrm{H}_{2} \mathrm{SO}_{4}=16 \mathrm{HMnO}_{4}+26 \mathrm{AsH}_{3}+130 \mathrm{CrS}_{3} \mathrm{O}_{1} 2+327 \mathrm{H}_{2} \mathrm{O}$
Algebra
Chapter 1
Linear Equations in Linear Algebra
Section 6
Applications of Linear Systems
Introduction to Matrices
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Okay, Um, so in these chemical reaction, we have six. Adam's here so that they are, um, McNamee's and suffer suffer. And arsenic and chromium and oxygen. Hydrogen. Okay, so for the first come down, we have one mechanism, one sober, and others will be zero second compound. We have zero magnetism, but we have, uh, excuse me. Zero suffer, but we have to our snake tin chromium, 35 oxygen and your hydrogen. All right, so for 1/3 compound, we have zero. I'm agonies one suffer zero zero for oxygen. Now we come to the right hand side of the equation. Remember? We need to put all the terms on the right inside to the defense eyes that we need to add a negative sign to all the John Alden entries on the right hand side here for the fourth compound. That is the 1st 1st campaign on right hand side. We have one agonies 00 zero and four oxygen and one hydrogen. Okay, since we are here, we're considering the Matrix. So at a negative sign to these three numbers. Now, for the fifth compound here, we have zero zero one. You new three. That is one arsenic and three hydrogen and add a negative sign here. 44. That's sick. Yeah, for the sixth compound. Here we have zero three, three silver and one chromium, 12 oxygen. We need to add a negative site and last compound we have. This is the water. So that's the deal with Jonesy. Role 01 two had, uh, making signs. All right, So here's our metrics. No, again, we usedto programming skills to find to find a reduced agent form. I'll just write down directly the reduce station here. So the the last of the NASA column for this reduce region of warm we have first for century will be next. 16 over three 27. Second entry is negative. 13 over 3 27 and third entry is connective 3 74 Over 3 27 The fourth entry will be active. 16 over 3 27 The faith one will be negative. 26 over 3 27 and the last entry here. Um, I have no room, but I will from the best here. How? Right down Just, uh, right next to the sentry. So that will be negative. 1 30 over 3 27 Okay, so this is our matrix. And we have one one and not that dog. Thio One and 00 on for the rest of entries here. Okay, so this Mitrice gives us then for Evan s. This compound will be 16 over 3 27 Off Wonder And this compound As to CR 10 0 35 will be 13 over 3 27 times water and suffer. Gassid will be sorry. We'll be three 74 over 3 27 off the water and for Hmm? 04 This will be six over three. 27 off the water. And but he asked as h three, this will be 26 over three. 27 up to water. And last you have CR as three old 12 is 1 30 over to be 27 off the water. Okay, now would take We take water as a 3 27 So? So our Kanko equation will be and my has We need We have a coefficient off 16. Plus here we have 13 for this compound. So that is 13 times they asked to cr 10. 0 35 plus So for the suffering that surfer casted, we have 3 74 The surfer gas it equals. Now here we have 16 or h hem and no whore. Thus, 26. Yes, it's three class 1 30 off CR as 3 12 See R s three old 12 US water water here will be 3 27 um three. 27 of water. And here is our solution.
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