Answer:
[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹
Explanation:
The balanced equation is
H₂O + Cl₂O ⇌ 2HOCl
Data:
Kc = 0.0900
[H₂O] = 0.004 32 mol·L⁻¹
[Cl₂O] = 0.004 32 mol
1. Set up an ICE table.
2. Calculate the equilibrium concentrations
![K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHOCl%5D%24%5E%7B2%7D%24%7D%7D%7B%5Ctext%7B%5BH%24_%7B2%7D%24O%5D%5BCl%24_2%24O%5D%7D%7D%20%3D%20%5Cdfrac%7B%282x%29%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%3D%200.0900%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B4x%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%26%3D%26%200.0900%5C%5C%20%5Cdfrac%7B2x%20%7D%7B0.00432%20-%20x%7D%20%26%20%3D%20%26%200.300%5C%5C2x%20%26%20%3D%20%26%200.300%280.00432%20-%20x%29%5C%5C2x%20%26%20%3D%20%26%200.001296%20-%200.300x%5C%5C2.300x%20%26%20%3D%20%26%200.001296%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B5.63%5Ctimes%2010%5E%7B-4%7D%7D%5C%5C%5Cend%7Barray%7D)
[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹
[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹
Check:
OK.
Answer:
(3rd option) Asexually reproduced offsprings have the same amount of genetic information as the parent, while sexually reproduced offsprings have half as much genetic information.
Explanation:
This is variation is caused as a result of both gamete production and fertilisation.
3.65. The mole ratio from the balanced equation is 2 moles CO2 : 2 moles CO. 2.
Answer:
We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution.
Explanation:
Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution. (MV) before dilution = (MV) after dilution. M before dilution = 1.0 M, V before dilution = ??? mL. M after dilution = 0.2 M, V after dilution = 100 mL. ∴ V before dilution = (MV) after dilution / M before dilution = (0.2 M)(100 mL) / (1.0 M) = 20.0 mL. So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution
