Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
<em></em>
Answer:
strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other.
Explanation:
In liquids, the attractive intermolecular forces are <u>strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other</u>.
Intermolecular forces are the forces of repulsion or attraction.
Intermolecular forces lie between atoms, molecules, or ions. Intramolecular forces are strong in comparison to these forces.
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Answer:
19.8 kg of C₂H₂ is needed
Explanation:
We solve this by a rule of three:
If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene
95.5×10⁴ kJ of heat may be released by the combustion of
(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂
Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g
If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg
The vapor pressure is obtained as 23.47 torr.
<h3>What is the vapor pressure?</h3>
Given that; p = x1p°
p = vapor pressure of the solution
x1 = mole fraction of the solvent
p° = vapor pressure of the pure solvent
Δp = p°(1 - x1)
Δp =x2p°
Δp = vapor pressure lowering
x2 = mole fraction of the of the solute
Number of moles of glycerol = 32.5 g/92 g/mol = 0.35 moles
Number of moles of water = 500.0 g/18 g/mol = 27.8 moles
Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles
Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012
Mole fraction of water = 27.8 moles/28.15 moles =0.99
Δp = 0.012 * 23.76 torr
Δp = 0.285 torr
p1 = p° - Δp
p1 = 23.76 torr - 0.285 torr
p1 = 23.47 torr
Learn more about vapor pressure:brainly.com/question/14718830
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Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
