Answer:
Approximately 10,5
Explanation:
The question is not really very specific, because it would need the percentages of those isotopes in the nature. As they are not shown, it should be the median of those two numbers.
atomic weight ≈ 
= 10,5
If you check a periodic table, you'll see it's actually 10,8, but that's because of the thing I told you at first (percentages missing).
Hope I could help.
<span>D. It shows that the electrons within an atom do not have sharp boundaries.</span>
Answer:
0.4 M
Explanation:
Molarity is defined as moles of solute, which in your case is sodium hydroxide,
NaOH
, divided by liters of solution.
molarity
=
moles of solute
liters of solution
Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,
mL
.
Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to
determine how many moles of sodium hydroxide you have in that many grams
convert the volume of the solution from milliliters to liters
So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.
7
g
⋅
1 mole NaOH
40.0
g
=
0.175 moles NaOH
The volume of the solution in liters will be
500
mL
⋅
1 L
1000
mL
=
0.5 L
Therefore, the molarity of the solution will be
c
=
n
V
c
=
0.175 moles
0.5 L
=
0.35 M
Rounded to one sig fig, the answer will be
c
=
0.4 M
Explanation:
Answer:
By the result of the formation of positive and negative ions, Ionic compounds are formed.
Explanation:
Electrons are actually transferred from one atom to another to form rare gas electron structures for each ion. The atom which forms a positive ion loses electrons to the atom which gains electrons to form a negative ion. A compound is not stable unless the number of electrons which are lost and gained are equal
They have the same density because a material, no matter how much of it there is, will always be a certain density. A 40g ball of iron has the same density as a 1g ball of iron.
