9.184 liters CH2O at STP
I think this is correct. Good luck
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
It takes 86 days take to cover half of the lake
Explanation:
In the day #1, the amount of the algae is X,
In the day #2 is 2X
In the day #3 is 2*2*X = X*2²
...
In the day #n the amount of the algae is X*2^(n-1)
Assuming X = 1m³. In the day 87, the area infected was:
1m³*2^(87-1)
7.74x10²⁵m³ is the total area of the lake
the half of this amount is 3.87x10²⁵m³
The time transcurred is:
3.87x10²⁵m³ = 1m³*2^(n-1)
Multiplying for 5 in each side:
ln (3.87x10²⁵) = ln (2^(n-1))
58.9175 = n-1 * 0.6931
85 = n-1
86 = n
<h3>It takes 86 days take to cover half of the lake</h3>
<span>The two techniques for separating an insoluble solid from a liquid are filtration and centrifuging. Filtration relies on the fact that the solid particles are smaller than the filter paper pores which allow the tiny molecules of liquid to pass through. With solids which dissolve in a liquid solvent, evaporation is commonly used to recover the solid and distillation is used to recover the solvent.</span>
The answer to letter b is
