Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>
It would be C, because Ionic bonds have to deal with valence electrons ( the outer shell ones)
Hi!
The correct options would be:
1. Cathode - <em>reduction</em>
The cathode is the negatively charged electrode, and so has an excess of electrons. Cations (positively charged ions) are attracted to the cathode, and gain electrons to acquire a neutral charge. The process in which a gain of electron occurs is called reduction.
2. Anode - <em>oxidation</em>
The opposite occurs at the anode which is positively charged and attracts negatively charged ions, anions. These anions lose their electrons at the anode to acquire a neutral charge, and the process involving loss of electrons is known as oxidation.
3. Salt Bridge - <em>ion transport </em>
Salt bridge is a physical connection between the the anodic and cathodic half cells in an electrochemical cell and is a pathway that facilitates the flow of ions back and forth these half cells. Salt bridge is involved in maintaining a neutral condition in the electrochemical cells, and its absence would result in the accumulation of positive charge in the anodic cell, and negative charge in the cathodic cell.
4. Wire - <em>electron transport </em>
Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.
Hope this helps!
Answer:
Strong acids are assumed 100% dissociated in water- True
As a solution becomes more basic, the pOH of the solution increases- false
The conjugate base of a weak acid is a strong base- true
The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion- True
As the Kb value for a base increases, base strength increases- true
The weaker the acid, the stronger the conjugate base- true
Explanation:
An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.
The pOH decreases as a solution becomes more basic (as OH^- concentration increases).
Ka refers to the dissociation of an acid HA into H3O^+ and A^-.
The greater the base dissociation constant, the greater the base strength.
The weaker an acid is, the stronger , its conjugate base will be.
Answer:
V= 12mL
Explanation:
you had the right idea with your Significant figures however, when we divide we see that it requires 2 significant figures as our least amount. this is because when looking at our division, 62 has 2 sig. fig. while 5.35 has a total 3. when looking at your answer we see that you had a total of 3 sig. figures. so in actuakity you had to round up to 12 and not to the tenths because the decimal makes .6 count as your third sig fig.
