87.
The density of the alloy is 105.0/10.12 = 10.3755
the density of Ag is 10.501
the density of copper is 8.933
so, in 1 gram of alloy, if we have x g of copper,
10.501(1-x) + 8.933x = 10.375
x = 0.080
So, the alloy is 8% copper
I'm using Zumdahl 7th Edition Chemistry Book and I'm having trouble with some practice problems in the first chapter.
Here are a few:
87: Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of 105.0 g and a volume of 10.12 m, calculate the mass percent of copper in the piece of necklace. Assume that the volume of silver present plus the volume of copper present equals the total volume.
Mass percent of copper =(mass of copper/total mass ) x100
93: At the Amundsen-scott south pole base station in Antarctica, when the temperature is -100.0 degrees F, researchers there can join "300 club" by stepping into 200.0F sauna then quickly run outside, around South Pole. What are those temps in C? [MY QUESTION: I tried to convert -100.0 F into Celcius, but the amount of significant figures I came up with were 4 (73.33), while the back of my book came up with 3 (73.3) In K? If temps only in C and K, can you become member of 300 club? (degree difference = 300)
11 answers
I don't quite understand why you multiplied 10.501 (density of silver) by (1-x). And what was the point of the problem talking about "assume that the volume of silver present plus the volume of copper.."?
Thanks for your help Steve
Thanks for your help Steve
Also it would be nice to see multiple ways, so do you think you could show me the other way in which the problem asks for the use of mass perecent of copper= (mass of copper/total) x100?
What's the volume? 10.12 m. What's m.
93. Google 300 club for a Wikipedia site to read about the 300 club. 300 club members have endured a change of 300 F (from 200 F to -100 F), To answer your question it is in F for the 300 degrees to be members of the 300 club. That is simply a matter of definition.
Do you have a question about 73.3?
93. Google 300 club for a Wikipedia site to read about the 300 club. 300 club members have endured a change of 300 F (from 200 F to -100 F), To answer your question it is in F for the 300 degrees to be members of the 300 club. That is simply a matter of definition.
Do you have a question about 73.3?
Yes. I understand the degree difference that they are asking for, but I'm having trouble with calculations. They give me the numbers 200.0 F and -100.0F and when I convert those to celcius, I think the significant numbers for each answer should have 4 significant answers. In the back of the book, I converted 200.0F correctly (93.33 C), but for -100.0F, I got 73.33 while the back got 73.33. I would like to know why?
Oh and sorry, 10.12 mL or 10.12 cm^3
With regard to 73.33 (note that's a -73.33) vs 73.3, I suspect the authors just goofed. Also note the correct spelling of celsius.
With regard to Steve's answer for the Cu/Ag necklace:
let x = grams Cu
Then 105.0-x = g Ag
Volume Cu + volume Ag = total volume
Then (mass Cu/d Cu) + (mass Ag/d Ag) = 10.12 cc. Then you solve for x = g Cu, plug that into your mass % formula to find mass % Cu. Steve just left out a step or two and assumed that since you were in AP chem you would work through the steps he gave you to find his answer and then all would be clear.
(The book tells you to assume the volume statement is true because it generally is NOT true) but it makes it a lot easier to work the problem assuming it is true.
With regard to Steve's answer for the Cu/Ag necklace:
let x = grams Cu
Then 105.0-x = g Ag
Volume Cu + volume Ag = total volume
Then (mass Cu/d Cu) + (mass Ag/d Ag) = 10.12 cc. Then you solve for x = g Cu, plug that into your mass % formula to find mass % Cu. Steve just left out a step or two and assumed that since you were in AP chem you would work through the steps he gave you to find his answer and then all would be clear.
(The book tells you to assume the volume statement is true because it generally is NOT true) but it makes it a lot easier to work the problem assuming it is true.
(Oh yeah, sorry about the sign, and I forgot a bracket after that perentices, I meant to say "what are the temperatures in Kelvin?".)
I'm sorry for my slowness, I'm actually learning from the textbook without any teacher guidance, so I can't ask a teacher except for online. Also, the answer in the book was 7.0% instead of 8%. Thank you both! I will try the process.
I'm sorry for my slowness, I'm actually learning from the textbook without any teacher guidance, so I can't ask a teacher except for online. Also, the answer in the book was 7.0% instead of 8%. Thank you both! I will try the process.
You're welcome to post here anytime. Look in your book for the density of Cu and the density of Ag. I suspect the numbers in your text are not quite the same as Steve used in his calculations. When posting problems like this it is helpful to post the numbers in your text; otherwise, our calculations are likely to be slightly different.
OH I GET IT NOW, the canceling of units This is great. Thank you again.
The numbers are different, so I will remember to post them next time.
1 answer