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Whhat ar 1 What V punt of the tbe FORMULA 1 carbon llowing reaction 1 Reky Entira Group reagent for the H 10.6 grams Fcmaln dloxide 1 otegoup HL complete carbon () ...

Question

Whhat ar 1 What V punt of the tbe FORMULA 1 carbon llowing reaction 1 Reky Entira Group reagent for the H 10.6 grams Fcmaln dloxide 1 otegoup HL complete carbon () [cact With : 1 grams of carbon 1 L

Whhat ar 1 What V punt of the tbe FORMULA 1 carbon llowing reaction 1 Reky Entira Group reagent for the H 10.6 grams Fcmaln dloxide 1 otegoup HL complete carbon () [cact With : 1 grams of carbon 1 L



Answers

How many grams of carbon molecules are in 1 mol of each substance?
a. methane, $\mathrm{CH}_{4}$
b. methanol, $\mathrm{CH}_{4} \mathrm{O}$
c. ethanol, $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$

The first item off. This question is very straightforward because the number off the mask off some atom in units off atomic mass is the same number as the molar mass off that atom. By that, I mean the following the molar mass off these atoms is 13.3 grams per mole. This is the number. So the atomic mass has the same number 13.3 but a different unit, a unit of atomic mass. So this is the mask off a single atom off 13 C. Now we have to find the mass off one atom off 13 C in kilograms. So all we have to do is convert these value two kilograms. We can do that by remembering that one unit off atomic mass has these number off kilograms. Then we have certain 0.3 times 1.6605 times stand toe minus 27 and these results in approximately true 0.1591 time stand to minus 26 kilograms.

The explanation of the solution is that for the party she right, the reaction has followed. That is stupefy her. The initial mosque off a is 1.60 Graham and decomposed for 38 minute the UN decomposed Mosque is 0.40 g. Uh, now we write the equation off half life for first order reaction is as follows that d t half is equal to 0.693 Divide by K hair D half is the half life and gazed, the rate constant. Mhm. Now we ride the rate off reaction for the first or the rest follows that are a is equal to minus K, and her Katie will be equal toe iron off 80 by a note, and we considered it as equation one. Now we substitute 1.60 grand for a note and 0.40 grand. Four a note and 30 minute for equation one as follows. This is for 80 and here they questioned it cake or toe one by 38 minute multiplied by on off 0.40 g, divided by 1.6 g, and the value off K is 0.364 minute universe does. The rate constant is zero point 0364 minutes. Universe. Now we calculate the half life. Yeah, as follows that he half is equal to 0.6 93 Divide by 0.364 minutes. Universe, By calculating this week at that 19, we're not Does the half life for the first order? Reaction is 19 minutes for the part B. Here we calculate the UN decomposed mass off a after one are as follows that Katie is equal toe iron off a off one Art by a note. Now here we put the values in the equation and solve this and we get the final values zero point 112 Now it's all for a one are as follows that a 41 artist, 0.112 multiplied by 1.6 g, which will be called to zero point 180 Graham does the UN decomposed mass off a after one. Art is 0.180 Graham. So this is the solution off the answer. Step by. Step in detail. Please go through this. Thank you.

Yeah, calculate the amount of carbon dioxide that could be produced to them. So we have first part, there is one mole of carbon is burned in here when one mole of carbon is embattled in here. Then we will get to in the balanced chemical action. We can observe that one mole of carbon was in one model of reaction to this. One more of kyoto. So quantity of kyoto will be one would yeah, in the B But we have One mole of carbon is banned in the 16 g oxygen. So first we have to calculate the whole of oxygen molecule two months by mother math. So mars of Is given 16 grand modern mass of photos are to grandpa removed And that will be equal to .5 more. Yes or no. So here now we can order that C plus for to give C. +02 So more of carbon is one mole of co. two is .5. So here or two is that limiting religion that will be finished initially. So more of sio two will be calculated by the help of photo so it has 0.5 mol ceo to yes or no And or we can say that 22 g e or two because if we want to calculate mass of kyoto then how can we calculate? We have mold of SEO to get these Children five months we have to calculate And mother most of you two is 44 g Permal Soma's ville here That is 22 grand. Okay, in the c part of the question we have two moles of carbon is bottled in 16 g of oxygen. So moral of auction that will comes out 0.5 mole of carbon is too Okay, so there will also or too is the limiting reagent And that's why I'm all of Kyoto will also 0.5 more if more of SEO to 0.5 than mars of Sio two weekend calculate. As have We calculated in the second part. That will be 22 grounds. Okay. Thanks.

He has so in this question were given a reaction. Where are reacting to A is forming products. This reaction is first order and were given that the initial mass at the start of the reaction is 1.6 programs. And after 30 minutes, this mass decomposes 2.40 grams in part a. Ross to find the half life and in part B were asked to find what mass of a is remaining after one hour. So the first thing I'm gonna do is I'm gonna find a relationship between the percent of a mass remaining and the initial percent of the mass without we're starting with. So I know we start with 100% of our initial mass, which is our initial, and that is 1.60 grams hips. And I know that my remaining is 0.4 grams. I'm gonna use these to find the percent remaining by putting my remaining over my initial mass and multiplying by 100%. And when I calculate that that is 25% remaining Now that I have this, I can use this value when I calculate my K, which is my rate constant And since I know this is first order, I'll use my first order integrated rate law equation, which is the natural log of my initial concentration. At some time, T is equal to the natural log of my initial concentration, minus Katie. I'm going to solve for Katie by adding that over to the left side and subtracting my natural log a sub t to the right side. And when I subtract a natural long over, I end up with this expression where my initial is going to be over my face empty, and this is due to my log, a rhythm rules solving for K. I get that K is equal to one over t times the natural log of my two concentrations. Now that I have this, I can go ahead and plug in my concentrations and my time value. So I want to find the time is at 38 minutes, which is given to us, and I'll be using the rest of what is given to me and the beginning of this question. So the natural log of my initial was, which is going to be 100% but I'll put that as a decimal so that would be just one. And my talk. My concentration at Time T, which is 38 minutes, is that 25% and that as a decimal 0.25 When I plug these values into the calculator, I find that my K value is equal to 0.365 in verse in minutes. Now that I have this, I can go ahead and use my half life equation to Seoul for my half life. So since this is first order, I can use my first order half life equation, which is equal 2.693 over K. And I can go ahead and plug in my K value that I just found, which is your 0.365 in verse minutes. And I find that my half life is equal to 19 minutes for this equation. Now that I have that, I can go ahead and go to Part B, where I'm going to use my integrated rate law to solve for the concentration that is, uh, needed or remaining Sizwe. So I have the natural log of my concentration at some time. Is he go to the natural log of my initial concentration minus Katie. I'm going to solve for the, um, concentration at Ace of Tea by first subtracting my initial natural law concentration over to the left side so I can combine my natural logs. When I do that, I have a similar relationship as last time, where it's the natural log over the two concentrations. This is going to be equal to negative, Katie. To get rid of the natural log, I'm going to exponentially eight both sides by it, putting both values to an exponent of E. Since natural log has a base of E. And when I do that, the ease cancel out and the natural law cancels out on the left side and I'm going to be left with my concentrations. And on the right side, I will have an exploded of e to the negative. Katie, I can go ahead and sell for a sub t. So a sub t is going to be equal to a subzero. He to the negative. Katie, I can go ahead and plug in my values. So my initial is 100% so that is just going to be one as a decimal and it's going to be e to the negative 0.365 in verse minutes, times t. So our T has to be one hour blood since we wanted in minutes. That's just going to be 60 minutes, just too much of our units and plugging this into our calculator, we find that the tea or the concentration is going to be equal to 0.112 And this is a percent since we have our initial percent of one. So if you just multiply this by 100% this is going to be equal to 11.2% of the initial. Since we want this as a mass, we're just going to go ahead and multiply this decimal by our initial mass. So we have our initial mass, we're gonna multiply it by the dust more that we found. So the mass remaining after one hour is going to be 0.185 grams


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