![ineedhelpireallydo](/avatars/8929.jpg)
ineedhelpireallydo
04.08.2019 •
Chemistry
Nickel (ni) has the fcc crystal structure, an atomic radius of 0.1246 nm, and an atomic weight of 58.69 g/mol. what is its theoretical density?
Solved
Show answers
More tips
- C Computers and Internet How to Check the Speed of My Internet?...
- H Health and Medicine 10 Ways to Cleanse Your Colon and Improve Your Health...
- W Work and Career How to Write a Resume That Catches the Employer s Attention?...
- C Computers and Internet Е-head: How it Simplifies Life for Users?...
- F Family and Home How to Choose the Best Diapers for Your Baby?...
- F Family and Home Parquet or laminate, which is better?...
- L Leisure and Entertainment How to Properly Wind Fishing Line onto a Reel?...
- L Leisure and Entertainment How to Make a Paper Boat in Simple Steps...
- T Travel and tourism Maldives Adventures: What is the Best Season to Visit the Luxurious Beaches?...
- H Health and Medicine Kinesiology: What is it and How Does it Work?...
Answers on questions: Chemistry
- C Chemistry 40 g of sodium reacts with g of chlorine to produce 117 g of sodium chloride *...
- C Chemistry Time of flight mass spectrometry Helpppppppppp Why is it -24 not -25...
- H History How did dyess’s account of the bataan death march differ from beck’s and burgos’s accounts...
- E English From bronx masquerade novel by nikki grimes what does tyrone s perspective add to the plot development? Explain. Pl, I really need help with this T^T...
- M Mathematics Help pleaseeeeeee...........
- E English Priest is to sermon as judge is to...
Ответ:
We will be calculating the density in units of gm/cm^3
Therefore, our standard units for this problem will be gram and cm
We are given that:
(1) Ni has an FCC structure, this means that: n = 4
(2) Ni has an atomic weight = 58.69 g/mol
(3) Ni has an atomic radius = 0.1246 nm
Now, the density of Ni can be calculated using the rule in the attached picture.
Examining the parameters in the rule, we will find that:
(1) n = 4
(2) A is the atomic weight of Ni = 58.69 g/mol
(3) Na is Avogadro's number = 6.022 * 10^23
(4) Vc is the number we want to calculate. We can find its value as follows:
Vc = a^3
a = 2R √2 where R is the radius = 0.1246 nm = 0.1256 * 10^-7 cm
Vc = (2R √2 )^3 = (√2 *2*0.1246*10^-7)^3
Vc = 4.377 * 10^-23
Now, we have all the parameters to substitute in the equation and get the theoretical density as follows:
theoretical density = [4*58.69] / [4.377 * 10^-23 * 6.022 * 10^23]
theoretical density = 8.9 gm / cm^3
Hope this helps :)
Ответ:
using the values of the periodic you first add the masses of C (12.01g) and O (there are two so it'll be 32.00g). That value will give the mass of 1mole of CO2.
I hate to do this, but
http://youtu.be/Pft2CASl0M0 is a link to a mr andersen video. I dislike watching these cause this is what my teacher uses instead of actually having to teach herself.