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:
Light
Explanation:
Plant is a thing so it has matter
Cheese is food so it has matter
Blood is a liquid so it has matter
Answer:
THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.
Explanation:
Mass = 3.0 kg = 3 * 1000 = 3000 g
Initial temperature = 10 C
Final temperature = 80 C
Change in temperature = 80 - 10 = 70 C
Specific heat of water = 4.18 J/g C
Heat needed = unknown
Heat is the amount of energy in joules needed to change a gram of water by 1 C.
Heat = mass * specific heat * change in temperature
Heat = 3000 g * 4.18 J/g C * 70 C
Heat = 877 800 Joules
Heat = 877.8 kJ.
The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.
Wave "B" is thinner. Is this the whole question?
