<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in =
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
Answer:
The mass of water to be added is 2 pounds
Explanation:
The given parameters are;
The mass of the given solution = 2 pounds
The concentration of the given solution = 30%
The desired concentration of the solution = 15%
The mass, m of the acetic acid in the given solution = 30% × 2 pounds
m = 30/100 × 2 pounds = 0.6 pounds
To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;
15% of X = 0.6 pounds
15/100 × X = 3/20 × 0.6 pounds
∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds
The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds
Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.
C because
Explanation Plato
The equation : y=3x-5
<h3>Further explanation
</h3>
Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates
General formula
y-y1 = m(x-x1)
or
y = mx + c
Where
m = straight-line gradient which is the slope of the line
x1, y1 = the Cartesian coordinate that is crossed by the line
c = constant
The formula for a gradient (m) between 2 points in a line
m = Δy / Δx

