Answer : The concentration of [CoCl₄]²⁻ is, ![2.95\times 10^{-3}M](https://tex.z-dn.net/?f=2.95%5Ctimes%2010%5E%7B-3%7DM)
Explanation :
Using Beer-Lambert's law :
![A=\epsilon \times C\times l](https://tex.z-dn.net/?f=A%3D%5Cepsilon%20%5Ctimes%20C%5Ctimes%20l)
where,
A = absorbance of solution = 0.935
C = concentration of solution = ?
l = path length = 0.71 cm
= molar absorptivity coefficient = ![446.9M^{-1}cm^{-1}](https://tex.z-dn.net/?f=446.9M%5E%7B-1%7Dcm%5E%7B-1%7D)
Now put all the given values in the above formula, we get:
![0.935=(446.9)\times C\times (0.71)](https://tex.z-dn.net/?f=0.935%3D%28446.9%29%5Ctimes%20C%5Ctimes%20%280.71%29)
![C=2.95\times 10^{-3}M](https://tex.z-dn.net/?f=C%3D2.95%5Ctimes%2010%5E%7B-3%7DM)
Therefore, the concentration of [CoCl₄]²⁻ is, ![2.95\times 10^{-3}M](https://tex.z-dn.net/?f=2.95%5Ctimes%2010%5E%7B-3%7DM)
Answer:
- 12.1
Explanation:
M(C3H8O3) = 3*12.0 + 8*1.0 + 3*16.0 = 92 g/mol
210.0 g C3H8O3 * 1 mol C3H8O3/92 g C3H8O3 = 210/92 mol C3H8O3
350.g = 0.350 kg H2O
Molality = mol soluty/ kg solvent = (210/92 mol C3H8O3) /(0.350 kg H2O) = =6. 522 molal
ΔT =i* Kf* m
(T2 - T1) = i* Kf* m
(T2 - 0°C) = 1*(-1.86°C/molal *6.522 molal)
T2= - 12.1°C
1) Equlibrium reaction
CH3COOH (aq) = CH3COO(-) (aq) + H(+) (aq)
2) Equilibrium constant
Keq = Ka = [CH3COO-] [H+] / [CH3COOH]
3) Equilibrium concentrations
CH3COOH CH3COO- H+
start 1.40 0 0
react x 0 0
produced 0 x x
equilibrium 1.40 - x x x
=> Ka = x * x / (1.40 - x)
Approximation: given that Ka is very small x <<< 1,40 and 1.40 - x ≈ 1.40
=> Ka ≈ x^2 / 1.40
=> x^2 ≈ 1.40Ka = 1.40 * 1.8 * 10^ - 5 = 2.52 * 10^-5
=> x ≈ √(2.52 * 10^-5) ≈ 5.02 * 10^ -3 M
4) pH = log 1 / [H+]
[H+] = x = 5.02 * 10^-3M
=> pH ≈ log (1 / 5.02 * 10^-3) ≈ 2.3
Answer: 2.3
Answer:
one cell to regulate
Explanation:
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Answer:
4.3 g
Explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 60.05 32.00
CH₃CH₂OH + 3O₂ ⟶ 2CO₂ + 3H₂O
Mass/g: 7.0
1. Calculate the moles of O₂
Moles O₂ = 7.0 × 1/32.00
Moles O₂ = 0.219 mol O₂
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2. Calculate the moles of CH₃CH₂OH
The molar ratio is 1 mol CH₃CH₂OH:3 mol O₂
Moles of CH₃CH₂OH = 0.219 × 1/3
Moles of CH₃CH₂OH = 0.0729 mol CH₃CH₂OH
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3. Calculate the mass of CH₃CH₂OH
Mass of CH₃CH₂OH = 0.0729 × 60.05
Mass of CH₃CH₂OH = 4.4 g