The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
Answer:
a producer and consumer relationship how several food chains and related.
O2- stands for superoxide anion, S2- stands for sulfide anion, H- stands for hydrogen anion, and Na+ stands for sodium this all ions possess the electron configuration of noble gas because each of them (O2-, S2-, H-, and Na+) has a full outer shell.
Answer:
It is Though the transfer of charges from one object to another, or (A).
Answer:
Y = 62.5%
Explanation:
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In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

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