Neutrons actually don't carry an electrical charge, which is why they are called neutrons because they are "Neutral".
Answer:
4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;
Explanation:
a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation
Answer:
CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)
Explanation:
When the calcium chloride dissolves. the calcium and chloride ions leave the surface of the solid and go into solution as hydrated ions.
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
Answer:
a.
b.
Explanation:
a. First, we solve the specific heat equation as follows:

b. Then, we use the molar mass of titanium to determine its molar heat capacity, as follows:
