Nickel metal will react spontaneously with the aqueous lead solution [Pb2+(aq)].
Oxidation is a chemical reaction involving the loss of electrons. In the reactivity series of metals, chromium, barium, and lithium are more reactive or more easily oxidized than nickel.
In the given problem, because nickel is in its elemental (solid metal) form, for it to react with the solution, it will have to be converted into its cationic form (which is positively-charged) by losing electrons. However, if the solution that nickel is reacted with is more easily oxidized compared to nickel, then the reaction will not be spontaneous because nickel is stable in this case.
On the other hand, if nickel is immersed in a solution of a metal which is more stable than it is, such as lead, nickel will be oxidized and lose its electrons to the more stable metal. The more stable metal in solution will accept the lost electrons of nickel and form a solid deposit, while nickel will be in its ionic form in solution.
Below is a reaction between nickel metal and aqueous lead nitrate solution.
16 is C
17 is E
18 is B
19 is A
20 is D
hope this helps :)
To balance the give n equation, we apply elemtal balance and count each elements per side. There are 2 nitrogens in the left side so there should be 2 moles of NO2. Since there are already 4 moles of O in the right side, there should be 2 moles of O2. Hence answer is a. <span>Place the coefficient 2 in front of oxygen and nitrogen dioxide.</span>
Answer:
See explanation below
Explanation:
In this case, HCl is a strong acid, therefore, it dissociates completely in solution.
To know the quantity of water we need to add, we first need to know the concentration of the acid with pH = 6:
[H⁺] = antlog(-pH)
[H⁺] = antlog(-6) = 1x10⁻⁶ M
This means that the concentration is being diluted.
Now, even if we add great quantities of water, and the concentration and volume change, there is one time that do not change despite the quantity of water added; this is the moles. So, all we have to do, is calculate the moles of the acid in the 1 mL of water, and then, the volume of the acid when it's dilluted:
moles HCl = 0.1 * (1/1000) = 1x10⁻⁴ moles
Now that we have the moles, we can calculate the volume which the acid with the lowest concentration has:
V = mol/M
V = 1x10⁻⁴ / 1x10⁻⁶
V = 100 L
This means that we need to add 99.999 mL of water