Gaseous methylamine, CH3NH2, can be classified as __________.(a) a Lewis base only(b) Arrhenius and Brønsted-Lowry bases(c) a Br
1 answer:
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Answer:
1. 0.97 V
2.
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ =
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL