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V₁/T₁ = V₂/T₂ = const
T₁=11+273=284K
V₂=3V₁
V₁/T₁=3V₁/T₂
T₂=3T₁
T₂=3*284=852K
t₂=852-273=579°C
<span>pH scale is used to determine how acidic, basic or neutral a solution is.
pH can be calculated using the Hydrogen ion concentration.
pH is calculated using the following formula
pH = -log [H</span>⁺<span>]
[H</span>⁺<span>] can be calculated knowing the pH
[H</span>⁺<span>] = antilog(-pH)
[H</span>⁺<span>] = 1.32 x 10</span>⁻¹⁰<span> M</span>
Answer:
230
Explanation:
Given equilibrium conditions, if a system is at equilibrium, this means the rate of a forward reaction becomes equal to the rate of a reverse reaction and the concentrations of each species become constant (stop to change).
This is defined by the equilibrium constant which defines the concentrations at equilibrium. For the given reaction, the equilibrium constant can be described as the ratio between the concentrations of products (raised to the power of their coefficients) and the concentrations of reactants (raised to the power of their coefficients):
![K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Q, on the other hand, is a reaction quotient. It is used only to determine where the equilibrium will shift when the system is actually not at equilibrium. The expression or Q is exactly same, but concentrations used would be non-equilibrium concentrations. There are three cases:
: in this case, equilibrium shifts towards the formation of products;
: the system is at equilibrium;
: equilibrium shifts towards the formation of reactants.
Looking at the context of our problem, since this system is at equilibrium, then Q = K = 230.
Answer:
1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.
2. Rainwater or Damp/moist air
3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws
Explanation:
1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.
In order for galvanic corrosion to occur, three elements are required.
i. Two metals with different corrosion potentials (anode and cathode)
ii. Direct metal-to-metal electrical contact
iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.
For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.
2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.
3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.
All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.