Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

Best regards.
Answer:
it'a answer number 2
Explanation: you divide the mass by volume and 32.2 divided by 4 is 8.05
Answer:
magnitude means absolute value, so the one that is greastest, like |-7| and |4| even id |-7| is a negative number, but it is still the one farthest away from 0, so |-7| is greater than |4|.
That is the way to find the greatest magnitude, but because I don't know your numbers so I can not answer your question, but this is the way to solve for it.
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When HCl is added to a saturated solution of PbCl2, the solubility of PbCl2 would decrease so precipitation would occur. The decrease in the solubility is due to the common ion effect or the presence of Cl- ions in both compounds.
There are
4.517
⋅
10
23
atoms of Zn in 0.750 mols of Zn.