0940j* 7Ak Uecimal numhcr ft !ic Wueink OE 4ud 92 00. 144 446 18*60 Nlt expre $ the unswer Complete the following- and [Hc nruch Ilber oL signilcunt lgung499 "...

Liquid iron at $1600 .^{\circ} \mathrm{C}$ has a density of $6.88 \mathrm{~g} / \mathrm{cm}^{3},$ and at $20 .{ }^{\circ} \mathrm{C}$ solid iron has a density of $7.86 \mathrm{~g} / \mathrm{cm}^{3} .$ Using these data, calculate the molar volume of iron at each temperature. Compare your calculated value with those in Table $12-1$ for different compounds.

To calculate the atomic mass of an element. We need to know what isotopes exist. For that element, the mass of each atom of the mass of the ice, that the mass of the isotopes, the atoms of the isotopes for each type of isotope and the percent natural abundance of each isotope than what we do is we take the mass of each isotope, multiplied by its fractional abundance, which which is just it's percentage Divided by 100. And we sum that up for each isotope. So for iron we have iron 54 With the mass of this isotope multiplied by its fractional abundance. 5.82% is .0582 Plus the mass of the second isotope multiplied by its fractional abundance, 91.66% is .9166 plus The third isotope mass multiplied by its fractional abundance plus the fourth isotope mass multiplied by its fractional abundance and we get 55.847 am you is the atomic mass of iron.

Using the ice topic information for iron. Let's calculate the average atomic mass Of iron. We'll need to take each individual isotopes. So the first isotope has a mass of 53.93 96 atomic mass units multiplied by the percent abundance but will convert the percent abundance to a decimal 58% is 580.0 5 8. And then the second isotope amounts of 55.9349. Yeah. And the percent abundance .9172 converted to decimal three days of topaz mass of 56.9354 and be .02-0 fourth isotope is 3:55, 9333 And% .0028. Solving here for the average atomic mass of iron, we get 55 79 96 Atomic Mass Units.

Given various isotopes, we should be able to determine the average atomic mass. But before doing that, This question wants us to complete a table similar to one found in the chapter. Where for each isotope, you need to write the symbol number of protons, number of neutrons, mass number and percent abundance. If we have iron 54, iron has an atomic number of 26. So we can write iron 50 for this way or this way. Iron 56. Very similar This way or this way. You'll notice that the atomic number 26 is the same down below. The mass number differs due to a different number of neutrons. I am 57 And Iron 58. So all of them are iron. So they have the same number of protons 26. Their mass numbers are provided 54, 56 57 and 58. Their mass numbers being the sum of protons and neutrons. So to figure out the number of neutrons, we take 54 -26 And we get 28 56 -26, 30 and then same thing for the last 2 31 and 32 and the percent abundances are provided in the problem. So I'm just rewriting them here to calculate the average atomic mass. I think it's easiest to take the percentage written as a decimal number. Remember to get a decimal number from a percentage. We're going to move the decimal three place or two places to the right one to gives us .0585 multiply that by the atomic mass of that particular isotope which is provided in the problem and then do the same thing for all four of the isotopes Percentage written as a decimal, multiplied by the mass percentages of decimal, multiplied by mass percentages. The decimal multiplied by mass, sum it all up and we get 55.845. Um you as the average weighted average atomic mass of iron. Mhm

Without any detailed calculations. Which one of the following will have the greatest mass of the element iron, the 1 kg pile of pure iron. We'll have the greatest mass since it has 100% iron by mass, and its unit of kilograms is larger compared to the rest, which would be B C nd, in which the mouse will come out to be in terms of grams and the percentage of iron in B, C and D is also less than 100%. So the greatest mass of iron is in a a 1 kg pile of pure iron filings.