how many ions of Na+ and Cl- will be present in 30g of nacl
Answers
Answer:
Here is what I get:
Explanation:
14.5g of
N
a
C
l
is equal to 0.248 mol according to the equation
n
=
m
M
Now, this equates to
1.495
×
10
23
Sodium chloride molecules.
We get this if we multiply 0.248 mol with Avogadro's number,
6.022
×
10
23
Each molecule of Sodium chroride holds two atomic ions bonded in an ionic bond-
N
a
+
and
C
l
−
in a 1:1 ratio. This means that the amount of molecules correspond with each of the individual ions.
So :
Number of Sodium ions=Number of Chloride ions=
1.495
×
10
23
ions
So the total amount of ions is twice this number,
2.990
×
10
23
ions in total.
Explanation:
14.5g of
N
a
C
l
is equal to 0.248 mol according to the equation
n
=
m
M
Now, this equates to
1.495
×
10
23
Sodium chloride molecules.
We get this if we multiply 0.248 mol with Avogadro's number,
6.022
×
10
23
Each molecule of Sodium chroride holds two atomic ions bonded in an ionic bond-
N
a
+
and
C
l
−
in a 1:1 ratio. This means that the amount of molecules correspond with each of the individual ions.
So :
Number of Sodium ions=Number of Chloride ions=
1.495
×
10
23
ions
So the total amount of ions is twice this number,
2.990
×
10
23
ions in total.
Explanation:
Answer:
Here is what I get:
Explanation:
14.5g of
N
a
C
l
is equal to 0.248 mol according to the equation
n
=
m
M
Now, this equates to
1.495
×
10
23
Sodium chloride molecules.
We get this if we multiply 0.248 mol with Avogadro's number,
6.022
×
10
23
Each molecule of Sodium chroride holds two atomic ions bonded in an ionic bond-
N
a
+
and
C
l
−
in a 1:1 ratio. This means that the amount of molecules correspond with each of the individual ions.
So :
Number of Sodium ions=Number of Chloride ions=
1.495
×
10
23
ions
So the total amount of ions is twice this number,
2.990
×
10
23
ions in total.
Explanation:
3.08x10~23
how
30g Nacl=.513moles
.513moles of Nacl contain 3.08x10~23 formula unit in ionic bonds
so Na+ions=30.8x10~23
Cl-ions=same as Na ions