Answer:
The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
Explanation:
For the complete recrystalization,
the amount of hot water should be such that, the benzoic acid is completely soluble in it.
As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.
we conclude that,
As, 68.0 grams of benzoic acid soluble in 1 L of water.
So, 0.700 grams of benzoic acid soluble in of water.
The volume of water needed = 0.01029 L = 10.29 mL
conversion used : (1 L = 1000 mL)
Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
<span>PbO
Let's look at each of the 4 compounds and see what's needed.
PbO.
* Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice.
PbCl4
* Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice.
Pb2O
* Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice.
Pb2S
* Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>
As the number of electrons added to the same principal energy level increases, atomic size generally
C. Decreases.
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
NH₂-CH₂-COOH + HNH-CH₂-COOH → NH₂-CH₂-CO-NH-CH₂-COOH + H₂O
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