Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺ | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺ + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺ + 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺ + 2e⁻⇄ Cu²⁺ + 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺ ⇄ Cu²⁺ + 2Ag
Answer:
They are held together by hydrogen bonds
Explanation:
Hydrogen bonds are special dipole-dipole attractions between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom(oxygen or nitorgen or fluorine).
Such molecules includes water, alkanoic acids, ammonia and amines.
A hydrogen nucleus has a high concentration of positive charge. The bond is actually an electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom(O or N or F).
Hydrogen bonds are very effective in binding molecules into larger units. Most substances that joins with hydrogen bonds have a higher boiling point and lower volatility.
This is why we have a strong intermolecular bond between water molecules.
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
Answer:
To calculate molarity, divide the number of moles of solute by the volume of the solution in liters. If you don't know the number of moles of solute but you know the mass, start by finding the molar mass of the solute, which is equal to all of the molar masses of each element in the solution added together.
Explanation:
try starting with 35.0 and dived it by the volume
