Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
<em>Q = 1.08x10⁻¹⁰</em>
As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate
Answer: remains constant.
Justification:
1) The phase changes are:
i) Boiling: pass from liquid to gas (absorbs heat energy)
ii) Condensation: pass from gas to liquid (release heat energy)
iii) Melting: pass from solid to liquid (absorb heat energy)
iv) Freezing: pass from liquid to solid (release heat energy)
v) Sublimation: pass from solid to gas (absorbs heat energy)
vii) Deposition: pass from gas to solid (release heat energy)
2) When a phase change occurs, whichever it is, the heat energy related with the process, either absortion or release, is used, to overcome the intermolecular forces (in the case of heat energy absortion) or to create stronger intermolecular forces (in the case of heat energy release).
Because of that, the heat energy exchange does not change the temperature of the substance.
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:
According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.
Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
Learn more: brainly.com/question/24245395
So boron has an electronic configuration of 1S2 2S2 2P1 and the element with an atomic number of 3 is Lithium which has an electronic configuration of 1S2 2S1.
So the answer is 1S2 2S1.
Answer:
29.75 Kg of NH₃
Solution:
In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,
6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂
Solving for X,
X = (5250 g × 28 g) ÷ 6 g
X = 25433 g of N₂
Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃
Solving for X,
X = (5250 g × 34 g) ÷ 6 g
X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃
