1L = 33.814 oz
xL = 2.75 oz
so it's a proportion
1L / 33.814 oz = xL / 2.75
solve for x
(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.
the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.
your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.
so with 3 significant figures,
0.0813272609 rounds to
0.0813 L
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(C) both are correct because if you are working in a Physical or health hazard area or if you are already working and you are going to be introduced to a hazardous situation you must be trained according to OSHA law.
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂
Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction
Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
Answer:
temp. of water, form of sugar
Explanation: