Resources found in lithosphere: gold and iron etc
Resources found in atmosphere: Water vapor, gases etc.
Explanation:
Sorry, I don't know, but I can tell you that when an atom, or a body, has the same amount of positive charges (protons) and negative charges (electrons), it is said to be electrically neutral. ... The net charge corresponds to the algebraic sum of all the charges that a body possesses.
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
<span>Ionization energy (IE) is the amount of energy required to remove an electron.
If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).</span>
Answer:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.03901 mol/liter
- [Cl₂O] = 0.02351 mol/liter
Explanation:
<u />
<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
<u />
<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)
<u />
<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
