It’s C.valley formed from water erosion
<em>n</em> = 15. A Bohr orbit with <em>n</em> = 15 comes closest to having a 24 nm diameter
.
The formula for the radius <em>r</em> of the <em>n</em>th orbital of a hydrogen atom is
<em>r</em> = <em>n</em>^2·<em>a</em>
where
<em>a</em> = the Bohr radius = 0.0529 nm
We can solve this equation to get
<em>n</em> = √ (<em>r</em>/<em>a</em>)
If <em>d</em> = 24 nm, <em>r</em> = 12 nm.
∴ <em>n</em> = √(12 nm/0.0529 nm) = √227 = 15.1
<em>n</em> must be an integer, so <em>n</em> = 15.
The electons are in the atmosphere
Answer:
3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr
Explanation:
The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.
Here was the process:
CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr
Balances PO4 (2on both sides)
CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Lithiums (6 on each side)
3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)
Hope this helped!
