Beryllium is a chemical element with symbol Be and atomic number 4. It is a relatively rare element in the universe, usually occurring as a product of the spallation of larger atomic nuclei that have collided with cosmic rays.
Answer:
The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.
The time needed for x oil changes is 12 * x.
The time needed for y tire changes is 18 * y.
The total time is the sum of the above times and needs to be less than 180 that is
12 * x + 18 * y < 180 divide both sides of equation by 6
12/6 * x + 18/6*y < 180/6
2*x + 3*y < 30
2*x < 30 - 3*y divide both sides by 2 to get the inequality for x
x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15
2*x + 3*y < 30
3*y < 30 - 2*x divide both sides by 3 to get the inequality for y
y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10
Also we can write x + y < x+ 3/2 * y < 15.
Explanation:
Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.
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⁻
