Answer:
Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.
Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ *
= 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
Ionic bonds are formed when one of the two atoms that are reacting has excess electrons and transfer the electrons to the atom that is deficient in electrons. During the formation of the ionic bond, one of the reacting atoms will donate electrons and form positive ion.
Answer:
The direction will be changed
.
Explanation:
According to the principles of motion, if an object is moving with solitary velocity and if an external force is applied to it, then that force changes the position, speed, and direction of that object.
In the given question, an external force is being applied to the toy which will change the direction of that toy.
Answer: The concentration of 
ions in vinegar is 0.001 M.
Explanation:
Given: pH = 3.0
pH is the negative logarithm of concentration of hydrogen ion.
The expression for pH is as follows.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the value into above expression as follows.
![pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C3.0%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%20antilog%20%28-%203.0%29%5C%5C%3D%200.001%20M)
Thus, we can conclude that the concentration of ions in vinegar is 0.001 M.
