Explanation:
Moles of compound =
We have ;
Volume of solution = 600 mL = 0.600 L ( 1 mL = 0.001 L)
Moles of NaOH = n
Molarity of the solution = 3 M
n = 3 M × 0.600 L = 1.800 mol
Mass of 1.800 mole sof NaOH :
1.800 mol × 40 g/mol = 72.0 g
Preparation:
Weight 72.0 grams of sodium hydroxide and add it to the 500 mL of volumetric flask along with some water. Dissolve the all the solute by adding small proportion of water. After the solution becomes clear make the water upto the mark of 500 ml.
Transfer the solution to a bigger beaker and 100 mL of water more to it.
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Answer:
7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.
Explanation:
Let the volume of 10% acid solution used to make the mixture = x L
So, the volume of 30% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 30 L .................. (1)
</u>
For 10% acid solution:
C₁ = 10% , V₁ = x L
For 30% acid solution :
C₂ = 30% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 25% , V₃ = 30 L
Using
C₁V₁ + C₂V₂ = C₃V₃
10×x + 30×y = 25×30
So,
<u>x + 3y = 75 .................. (2)
</u>
Solving 1 and 2 we get,
<u>x = 7.5 L
</u>
<u>y = 22.5 L</u>
Answer:
A liquid
Explanation:
This is the answer!! Hope this helps.
