Explanation:
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
A. A design is remodeled after analysis and tested again.
Answer:
The section of the bar is 2.92 inches.
Explanation:
Mass of the steel cut ,m = 1.00 kg = 1000 g
Volume of the steel bar = V = Area × height
Height of the of the section of bar = h
Area of Equilateral triangular = ![\frac{\sqrt{3}}{4}a^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7Da%5E2)
a = 2.50 inches
Cross sectional area of the steel mass = A
![A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%282.50%20inches%29%5E2%3D2.71%20inches%5E2)
![V = 2.71 inches^2\times h](https://tex.z-dn.net/?f=V%20%3D%202.71%20inches%5E2%5Ctimes%20h)
Density of the steel = d =![7.70 g/cm^3](https://tex.z-dn.net/?f=7.70%20g%2Fcm%5E3)
![1cm^3 = 0.0610237 inches^3](https://tex.z-dn.net/?f=1cm%5E3%20%3D%200.0610237%20inches%5E3)
![d=\frac{m}{v}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7Bm%7D%7Bv%7D)
![\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}](https://tex.z-dn.net/?f=%5Cfrac%7B7.70%20g%7D%7B0.0610237%20inches%5E3%7D%3D%5Cfrac%7B%201000%20g%7D%7B2.71%20inches%5E2%5Ctimes%20h%7D)
![h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B%201000%20g%5Ctimes%200.0610237%20inches%5E3%7D%7B2.71%20inches%5E2%5Ctimes%207.70%20g%7D)
h = 2.92 inches
The section of the bar is 2.92 inches.
In order to write the Aluminium electron configuration we first need to know the number of electrons for the Al atom (there are 13 electrons). When we write the configuration we'll put all 13 electrons in orbitals around the nucleus of the Aluminium atom.
Answer:
always repeats the same unit
made of smaller molecules
often made from repeating units