Answer:
C
Explanation:
Large chlorine atoms can not fit within the atoms of boron
Answer:
The atomic number of<span>N<span>157</span></span>
The number of protons is 7
The number of electrons is 7
The number of neutrons is 8
Explanation:
The atomic number of Nitrogen is 7 because Nitrogen has 7 protons.
The seven protons attract 7 electrons in the ground state.
If the atom had fewer or more than 7 protons the atom would not be Nitrogen.
The mass of the atom is the sum of protons and neutron. so
p + n = mass ( protons (p) and neutrons(n) both have an atomic mass of one
7 + n = 15 subtract 7 from both sides
<span>7−7+n=15−7</span>
n = 8
even tho the guy copied the answer its true neon can form a compound
Answer:
The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.
Explanation:
Answer:
The answer is:
(a) 
(b) NaCl
(c) 0.211 g
Explanation:
Given:
The mass of NaCl,
= 0.0860 g
The molar mass of NaCl,
= 58.44 g/mol
The volume of
,
= 30.0 ml
or,
= 0.030 L
Molarity of
,
= 0.050 M
Moles of NaCl will be:
= 
= 
= 
now,
Moles of
will be:



(a)
The reaction is:
⇒ 
(b)
1 mole of NaCl react with,
= 1 mol of 
0.0015 mol
needs,
= 
Available mol of NaCl < needed amount of NaCl
So,
The limiting reagent is "NaCl".
(c)
The precipitate formed,
= 
= 