Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
The answer to this question would be C. I hope this helps!
Answer:
NaCl
Explanation:
Thus, for the compound between Na + and Cl −, we have the ionic formula NaCl (Figure 3.5 "NaCl = Table Salt").
Answer:
6.0 L
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 0.075 M
V1 = 200 L
M2 = 2.5 M
V2 = ?
Solve for V2 --> V2 = M1V1/M2
V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L
Answer:
Molarity is halved when the volume of solvent is doubled.
Explanation:
Using the dilution equation (volume 1)(molarity 1)=(volume 2)(molarity 2), we can demonstrate the effects of doubling volume.
Suppose the starting volume is 1 L and the starting molarity is 1 M, and doubling the volume would make the final volume 2 L.
Plugging these numbers into the equation, we can figure out the final molarity.
(1 L)(1 M)=(2 L)(X M)
X M= (1 L x 1 M)/(2 L)
X M= 1/2 M
This shows that the molarity is halved when the volume of solvent is doubled.
