Your answer is C for sure
Answer:
See explanation.
Explanation:
Hello,
In this case, we could have two possible solutions:
A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:
That is the mass of copper (II) sulfate contained in 1 mol of substance.
B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:
So you can solve for the moles of the solute:
Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:
But this is just a supposition.
Regards.
The radius of the cation is much smaller than the corresponding neutral atom.(b) The radius of an anion is much larger than the corresponding neutral atom.Explanation:The size of the atom or ion is inversely proportional to the nuclear charge experienced by the electrons.(a)The size of the cation is smaller than the size of the corresponding neutral atom. This is because after removal of an electron from the highest principle energy level the nuclear charge experienced by the valence electrons increases resulting in the decrease in size.(b)The size of an anion is larger than the size of the corresponding neutral atom. In an anion, an extra electron is added to the highest principle energy level but the effective nuclear charge pulling the electrons towards the nucleus is still same. The net effective nuclear charge experienced by the electrons present in the outermost shell decrease. Moreover, due to the added electron, the repulsion between the electrons also increases resulting in the increase in size
Make since? i hope this helps
21 kg x [(3 x 35.45)/(12.01 + 19.00 + (3 x 35.45))] =
21 kg x (106.35/137.36) = 16.3 kg of chlorine
You just multiply the weight of the material by the fraction of chlorine (by weight). The others are done the same way
