Copper forms a green deposit on its surface when exposed to air which is usually a mixture of copper carbonate and copper hydroxide.
Ok so the PH for each solution will be
<span>1.) 1.638
2.) 5.2
3.) 12.524
This is found using the formula: </span> pH<span> = - log [H</span>3O+].
And the pOh for each one will be
<span>1.) 12.362
2.) 8.8
3.) 1.4763
This is found by the formula </span>pOH = -log10[OH-<span>]
</span>Hope this is useful
In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following
, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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It should be 3-4 dallors because 8x3 is 24 and that is 3 pounds and you only have one banana left so is supposed to be 3 dallors and change.
Answer:
4.7485 g
Explanation:
4.50 x 10^22 Cu atoms * (1 mol Cu / 6.022 x 10^23 Cu atoms) * 63.546 g Cu/(mol Cu) = 4.7485 g
In every mole of Cu, there are 6.022 x 10^23 atoms (Avogadro's number). The molecular weight of copper is 63.546 g/mol.