Answer:
The correct answer is 0, 235 mol
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:
1 atm x 5, 25l = n x 0, 082 l atm / K mol x 273 K
n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K
n= 0, 235 mol
For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.
C Linnaeus was the first person known to have used the terms genus and species when classifying organisms.
A) releasing CO2 that dissolves and forms acid in the oceans (i think i'm sorry if its wrong)
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
