Answer:
the 3rd one (0.01 cm the one selected already)
Explanation:
copper wire isn't excessively big, and it wraps around the pencil because its malleable. I think that the most accurate would be 0.01 cm
The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.
An ion can be cation (positively charged) or anion (negatively charged).
Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).
The remaining elements are completed as follows to attain octet structure;
<u>Element</u>--<u>valence electron</u>--<u>electrons to gain</u>--<u>electrons to lose</u>--<u>ion formed</u>
O ------------ 6 ---------------------- 2 ------------------------ none -------------- 
Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ 
Br ----------- 7 --------------------- 1 ------------------------ none --------------- 
S ------------ 6 ----------------------- 2 ------------------------ none --------------- 
Cl ------------ 7 ----------------------- 1 ------------------------ none ----------------
K -------------- 1 ----------------------- none ----------------------- 1 ------------------ 
Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- 
Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- 
Learn more here: brainly.com/question/21089350
It would emit energy in most of the cases in form of light
Before an article is published in a scientific journal or in any "peer-reviewed" journal the article is reviewed thoroughly by scholars from the journal as well as peers or scholars of the articles author from the same field. This process occurs to provide credibility to the ideas being published and so that readers and other scholars can rely on the validity of the material being published.
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M