What is the purpose of finding oxidation states in the half-reaction method for balancing equations?

Look at the data table and graph for the 25-coil electromagnet. With the 1.5 V battery, the

Margarita [4]

Answer a:
A 1.5 V battery, the electromagnet picked up an average of 6 paper clips, while with the 6.0 V battery, an average of 23 paper clips were picked up. Battery of 6.0V is 6.0/1.5 = 4 times stronger than battery of 1.5 V

Answer b:
Ratio of the number of paper clips picked up using the 6.0 V battery to the number picked up using the 1.5 V battery is = 23/6 = 3.8 ≈ 4.

Answer c:
As the voltage power increase, more paper clips were picked up by electromagnet. This indicated that there is a direct relationship. Mathematically it can be expressed as:

Voltage Power α Number of paper clips that were picked up

4 0

3 years ago

True or false: the most easily ionizable elements are the most electronegative.

ehidna [41]

False, This is because when you can easily ionize and atom or the chances of it being ionizable are quite high, it means that that particular atom have very low ionization potential that is the reason why it was easily ionizable An atom with a high ionization power and a firmly negative electron fondness will both pull in electrons from different particles and oppose having its electrons taken away; it will be an exceedingly electronegative molecule.

3 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

Examine the diagram shown. Which are the correct labels for continental plates 1 and 2?

Over [174]

Answer:

Australian and eurasian

Explanation:

8 0

3 years ago

Why does the rate of a reaction increase when the concentration of reactants is increased?

leva [86]

If reactants eventually collide, there is an occurrence of reaction.

Therefore, when there is an increase concentration of reactant, meaning to say that there are several moles of it every unit volume. An example of this is a room having hundred of people will absolutely get higher concentration compared to a room with one individual only.

Pertaining to effective collisions, if ever there is an increase of concentration, the frequency and rate of effective collisions among reactants surges in such a way that the rate of reaction also surges. Same with passing into a room with only 1 individual compared to hundred people blind persons, you probably want to proceed to the room with several people.

This is the simple logic behind that scientific existence.

5 0

3 years ago