Because you are getting rid of the acid in the juice so you are balancing the equilibrium because you have lesd acid and the base is the same making it more equal
Answer:
2L of water.
Explanation:
To know the volume of water to be added to the initial solution, first let us calculate the volume of the final solution. This is illustrated below:
Data obtained from the question:
Initial volume (V1) = 2L
Initial concentration (C1) = 6mol/L
Final concentration (C2) = 3mol/L
Final volume (V2) =?
Using the dilution formula, we can obtain the final volume of the stock as follow:
C1V1 =C2V2
6 x 2 = 3 x V2
Divide both side by 3
V2 = (6 x 2)/3
V2 = 4L.
The final volume of the solution is 4L.
To obtain the volume of water added, we shall determine the change in the volume of the solution. This is illustrated below:
Initial volume (V1) = 2L
Final volume (V2) = 4L
Change in volume = V2 – V1 = 4 – 2 = 2L.
Therefore, 2L of water must be added to the initial solution.
The density has been given as the mass of the substance per volume of the solution. The mass of citric acid in the bottle has been 1.22 grams.
<h3>What is percent mass?</h3>
Percent mass can be given as the mass of substance in the total mass of the solution.
The volume of the face serum has been 250 mL, and the density is 0.979 g/mL.
The mass of face serum in the bottle has been:

The total mass of face serum has been 244.75 grams.
The percent mass of citric acid in face serum is 0.5%. This states the presence of 0.5-gram citric acids in 100 grams sample.
The mass of citric acid in the face serum sample is given as:

The mass of citric acid present in 250 mL serum sample has been 1.22 grams.
Learn more about percent mass, here:
brainly.com/question/5394922
Answer:
R = V / I
Explanation:
Ohm's Law:
V = I * R , V = voltage , I = current , R = resistance
rearranging to solve R,
R = V / I
Answer
MnO₄ + 2H⁺ +3NO₂⁻ →3NO₃⁻ + Mn²⁺ +H₂O
Explanation
This is a redox reaction (oxidation-reduction reaction) which involves the transfer of electrons between two species. i.e
Mn + 6e⁻→Mn²⁺ (reduction)
3N³⁺- 6e⁻→3Mn⁵⁺(oxidation)