Answer:
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Explanation:
corncobpipe accident
The answer is Three
!!!!!!
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:
44.8 L of O2 will react (option D)
Explanation:
Step 1: Data given
Number of moles of SO2 = 4.00 moles
STP = Pressure = 1 atm and temperature = 273 K
Step 2: The balanced equation
2 SO2(g) + O2(g) → 2 SO3(g)
Step 3: Calculate moles of O2
For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3
For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2
Step 4: Calculate volume of O2
For 1 mol we have a volume of 22.4 L
V = (n*R*T)/ p
V = (2.00 * 0.08206 * 273)/p
V = 44.8 L
For 2.00 moles we have a volume of 2*22.4 = 44.8 L
44.8 L of O2 will react (option D)
Answer:
8.1433 g of XeF₆ are required.
Explanation:
Balanced chemical equation;
XeF₆ (s) + 3H₂ (g) → Xe (g) + 6HF (g)
Given data:
Volume of hydrogen = 0.579 L
Pressure = 4.46 atm
Temperature = 45 °C (45+273= 318 k)
Solution:
First of all we will calculate the moles of hydrogen
PV = nRT
n = PV/ RT
n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K
n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹
n = 0.0995 mol
Mass of hydrogen:
Mass = moles × molar mass
Mass = 0.0995 mol × 2.016 g/mol
Mass = 0.2006 g
Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.
H₂ : XeF₆
3 : 1
0.0995 : 1/3× 0.0995 = 0.0332 mol
Now we will calculate the mass of XeF₆.
Mass = moles × molar mass
Mass = 0.0332 mol × 245.28 g/mol
Mass = 8.1433 g
