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(e) 347 mg of acetaminophen, HOCsHANHCOCH3 compound mol mol : mol mol mol...

Question

(e) 347 mg of acetaminophen, HOCsHANHCOCH3 compound mol mol : mol mol mol

(e) 347 mg of acetaminophen, HOCsHANHCOCH3 compound mol mol : mol mol mol



Answers

Acetaminophen, an analgesic, has the molecular formula $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}$ (a) What is the molar mass of acetaminophen? (b) How many moles are present in $5.32 \mathrm{~g}$ acetaminophen? (c) How many grams are present in $0.166 \mathrm{~mol}$ acetaminophen?

Acetaminophen has the chemical formula c. eight. Each nine 02 N. For a rusty, calculate the molar mass of acetaminophen. She had eight carbons eat Times 12.0 g per mole. Give us 96 g per mole. Yeah. And then we have nine hydrogen. Mhm And clint zero grampa molly. So you combine grand promote. Yeah. To oxygen. I'm 16.0 grounds perm. All this gives us 32.0 grounds for more. We have one nitrogen one times 14.0 g per mole. Therefore our molar mass. Yeah. Or c the men often works out to 151 0 grams promote adding everything together. For B, calculate the amount of acetaminophen and 532 g. If we have 5.32 g of C. Eight age nine 02 N. You know them one more. He ate 902 n. has a mass of 151.0 g. Yeah. Before this would be equal to 00352 malls of C eight H902 10 paper. See Calculate the mass of a scene of benefit and .166 moles. We actually won. 6 6 moles, the c eight 9 to 10 Molar Mass 151.0 g. Mhm. In one more. Yeah, this would be equal to 25.1 grams of the sea Each 9 4- eight.

For this problem, we'll start with 5.12 molds aluminum to calculate grams aluminum. We simply multiply by the molar mass of aluminum, which from the periodic table is 26.98 g. So 5.12 moles corresponds to 138 g aluminum .75 moles of copper. When multiplied by its smaller mass, corresponds to 47.7g copper 3.52 moles of magnesium bromide, which has a molar mass. That will be the molar mass of or the massive one mole of magnesium plus the massive two moles bro. Mean, which is when 84.1 g. So 3.52 moles of magnesium bromide is 648 g magnesium bromide. For the next one, we have a compound C two H six. So, so that means its molar mass will be the mass of two moles of carbon plus six moles of hydrogen plus one mole of oxygen, which is 46.7 g. So 0.45 moles of the compound corresponds to 6.68 g, And the last one we have 2.08 moles of ammonium sulphate, ammonium sulfate. When calculating its smaller mass is a little more tricky, but you can still do it. It's two moles nitrogen Plus the massive eight moles hydrogen plus the mass of one mole sulfur, and formals oxygen gives us 132.14 g, so two point oh eight moles ammonium sulfate corresponds to 275 g ammonium sulfate.

To solve this problem, we will use the molar mass off each other. Exceptions given in the book One mole off aluminum is equal in tow, 26.98 g off aluminum, which is also called as the mamas off aluminum. Therefore, 3.57 months off. Aluminum will have 96.3 g off aluminum, using a single approach where silicon RC has 28.9 g. As the mullahs mass, we get the total number off grams in 42.6 miles off a sigh as equal toe 1.20 multiplied by 10 part 3 g off a sigh.

For this question asks, Determine the mass in grams for each of the following, but for letter ate, it asks for 3.57 moles of aluminum A l. So in order to do this, you need to look up the molar mass of aluminum. Every one mole of a L. I find a table. It has a more massive 26.92 But that millions that for every one mole of aluminum, it has a massive 26.982 I put that on the top so that being moles can cancel and we're left with grams. And then when I put that into my calculator, when I get is 96 granted. So that was for letter A for a letter B. It asks for 42.6 moles of Silicon Z s I. And what I do is I find the molar mass of silicon the one let's eight 0.86 grams, and that is on the periodic table. Moles canceled. I go ahead and plug that into my calculator. And when I get is 1196 point five grams, uh, silicon


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