When that happens, you get a plasma — the fourth state of matter.
Answer:
Anything not on the periodic table is an element non example! ... So, for a substance to be an element, all of its atoms must have the same number of protons. Examples of elements include hydrogen, lithium, nickel, and radium.
Explanation:
1mol—44g/mol
0.10mol—x
x=0.10*44
x=4.4 g
Answer:
Explanation:
Naming of the ionic compounds:-
- The name of the cation is written first and the the name of the anion is written after the name of the cation separated by single space.
- The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
-
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.
Hence, given ionic compound:-
Cobalt(II) phosphate
So, Cobalt will have a positive charge of +2
Phosphate is 
So, the formula is :-
Co
2 3
Answer:
- <em>The mystery substance is</em> <u>C. Bromine (Br) </u>
Explanation:
<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.
<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.
Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).
The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.
The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.
Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.
You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.
Therefore, <em>the mystery substance is bromine (Br).</em>
