In a solution that is 90% ethyl alcohol and 10% water, the solute is 10 g and the solvent is 90 g.
given that :
The solute in the solution by mass percent = 10 %
mass percent = ( mass of solute / mass of solution ) × 100%
mass of solute in the solution is = 10 g
The solvent present in solution by mass percent = 90 %
the mass of solvent in the solution is is 90 g
Thus, In a solution that is 90% ethyl alcohol and 10% water, the solute is 10 g and the solvent is 90 g.
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Answer:
A.All Mixtures Are Made Up Of Solutions
Explanation:
Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:

The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
The net ionic equation : Ag⁺(aq)+l⁻(aq) ⇒ Agl(s)
<h3>Further explanation</h3>
Given
Reaction
AgF(aq)+Nal(aq) ——> Agl(s)+NaF(aq)
Required
The net ionic equation
Solution
The equation of a chemical reaction can be expressed in the equation of the ions
In the ion equation, there is a spectator ion that is the ion that does not react because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
AgF(aq)+Nal(aq) ——> Agl(s)+NaF(aq)
Ag⁺(aq)+F⁻(aq)+Na⁺(aq)+l⁻(aq) ⇒ Agl(s)+Na⁺(aq) + F⁻(aq)
<em>Ag⁺(aq)+l⁻(aq) ⇒ Agl(s)</em>