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
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The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.
Given:
The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.
To find:
The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.
Solution:
- The maximum height attained by nitrogen gas molecule = h
- The temperature of nitrogen gas particle = T = 330 K
The average kinetic energy of the gas particles is given by:

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy
- The potential energy at height h =

- Molar mass of nitrogen gas = 28.0134 g/mol
- Mass of nitrogen gas molecule = m

- The acceleration due to gravity = g = 9.8 m/s^2
- The maximum height attained by nitrogen gas molecule = h
- The potential energy is given by:


The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.
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D.
both are stated to be in aqueous solutions by the (aq)
Its both A and C because both A and C have only one type each so it can only be those two :)