When a neutral hydrogen atom loses an electron, a positively-charged particle should remain.
Answer:
C. Lithium is most easily oxidized of the metals listed on the activity series and therefore it will most easily give electrons to metal cations
Explanation:
"Lithium" is a type of alkali metal that has a "single valence electron." Since it is a reactive element, it easily gives up an electron when it is combined with other elements. Such giving up of electron is meant to create compounds or bonds.
Among the common metals listed, "lithium" is the most easily oxidized. This means that it donates its electrons immediately. Such combination makes it exist as a<em> "cation"</em> or <em>"positively-charged."</em>
So, this explains the answer.
Density- The proportion of mass to volume of an object.
<span>Density determines what floats and what sinks. More dense sinks less dense will float.</span>
Answer:
The value of entropy change for the process 
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume 
= 4 L
Final volume 
= 6 L
Gas constant for this ideal gas ( R ) = 
Where 
= Universal gas constant = 8.314 
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 
Entropy change at constant temperature is given by,
Put all the values in above formula we get,
![dS = 0.0224 log _{e} [\frac{6}{4}]](https://tex.z-dn.net/?f=dS%20%3D%200.0224%20%20log%20_%7Be%7D%20%5B%5Cfrac%7B6%7D%7B4%7D%5D)
This is the value of entropy change for the process.
this is not my work
-Brooks Nelson
Brooks Nelson, Chemist at University of Florida
Answered Oct 12, 2018 · Author has 368 answers and 54.1k answer views
My limited understanding is you need pressure, temperature and enough elements that can fuse. If the temperature and pressure aren't high enough and/or you don't have enough elements that can fuse, then no fusion.
In fact I've never heard of fusion in a nebula, only in a star. The exception being a brown dwarf, which is considered substellar at 10 to 90 Jupiters in mass, and they can fuse deuterium (if over 13J) and also lithium (if over 60 J). But the burn through all of it in about 10 million years and wouldn't emit light like a main sequence star would.
