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:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
Answer : The vapor pressure of bromine at 
is 0.1448 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of bromine at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of bromine = 
= heat of vaporization = 30.91 kJ/mole = 30910 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of bromine at is 0.1448 atm.
Answer:
0.1g (Gallon) of chlorine
Explanation:
<u>Formula</u>
1 gallon = 3.7L; the density of water is 1.0g/ml
<u>Given</u>
2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water
<u>Solve</u>
If 2g (gallon) chlorine = 1,000,000g (gallon)
∴, ? chlorine = 40,000
The First step; set up an equation
1000000/2 = 40000/?
The Next step; divide 1 million to 2
1000000 ÷ 2 = 500000
Then, divide the result by 40000
40000 ÷ 500000 = 0.08
In the nearest unit that is 0.1
Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.
The movement of rock pieces and other materials on earth's surface is called Weathering
