You start by using proportions to find the number of liters of solution:
180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution
=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.
If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.
Answer: 0.1 liter of water.
Answer:
nah but ill take points tho
Explanation:
Answer:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.03901 mol/liter
- [Cl₂O] = 0.02351 mol/liter
Explanation:
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<u>1. Chemical reaction:</u>

<u>2. Initial concentrations:</u>
i) 1.3 g H₂O
- Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
- Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter
ii) 2.2 g Cl₂O
- Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
- Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter
<u>3. ICE (Initial, Change, Equilibrium) table</u>

I 0.0481 0.0326 0
C -x -x +x
E 0.0481-x 0.0326-x x
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<u>4. Equilibrium expression</u>
![K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cdfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D.%5BCl_2O%5D%7D)

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<u>5. Solve:</u>

Use the quadatic formula:

The positive result is x = 0.00909
Thus the concentrations are:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
- [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

The explanation is shown in the image attached with
Answer:
the oxidation number of those elements is 2 because some of them are molecules