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 correct answer would be C2H60
Adding hot cocoa to hot water would be a chemical change
Explanation:
The given data is as follows.
(NaCl) = 
(H-O=C-ONO) = 
(HCl) = 
Conductivity of monobasic acid is 
Concentration = 0.01 
Therefore, molar conductivity (
) of monobasic acid is calculated as follows.

= 
= 
= 
Also,
= 
= 
= 
Relation between degree of dissociation and molar conductivity is as follows.

= 
= 0.1254
Whereas relation between acid dissociation constant and degree of dissociation is as follows.
K = 
Putting the values into the above formula we get the following.
K = 
= 
= 
= 
Hence, the acid dissociation constant is
.
Also, relation between
and
is as follows.

= 
= 3.7454
Therefore, value of
is 3.7454.
Answer: The amount of energy needed to move an electron from one zone to another is a fixed, finite amount. The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level.
Explanation: