The cathode electrode is copper. This means that the concentration of copper must be increased to increase the cell potential. This can be done by diluting the Al3+ solution. Increasing the surface area will not change the current but will increase the voltage. The answer is (D) I and III only.
Answer:
The molecules absorb heat and acquire more kinetic energy.
Explanation:
In a solid, the solids only vibrate about their mean positions but do not translate. When energy is supplied to the molecule in the form of heat, the molecules vibrate faster. Eventually, they acquire sufficient energy to leave their mean positions and translate. Hence the solid crystal collapses.
When ice is heated, water molecules acquire sufficient kinetic energy to translate. The intermolecular bonds are gradually broken in the solid framework as heat is absorbed. The heat required for this is known as the latent heat of fusion.
The temperature remains constant until phase transition is over, then temperature rise resumes.
Answer:
Decomposition Reaction
Explanation:
If you are referring to what type of reaction that occurred then the answer would be decomposition reaction.
This is a chemical reaction where one reactant is broken down into two or more products.
REACTANT → PRODUCT
AB → A + B
The products can be two or more elements or two or more compounds, depending on what was decomposed.
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
