Answer: D. Slow down the chain reaction by absorbing free neutrons
Explanation: just got it right on the quiz A P E X
Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.
So first, we take our measured value, .299 cm, minus our known value, .225 cm.
.299 cm - .225 cm=.004 cm
Next, we divide that by our known value

Finally, multiply your answer by 100
.0177777778 x 100= 1.77777778 %
Round to three significant figures, and you're done.
=1.78 % error
Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S = 
Number of moles of S =
= 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
Answer:
2-methoxybutane
Explanation:
This reaction is an example of Nucleophilic substitution reaction. Also, the reaction of (S)-2-bromobutane with sodium methoxide in acetone, is bimolecular nucleophilic substitution (SN2). The reaction equation is given below.
(S)-2-bromobutane + sodium methoxide (in acetone) → 2-methoxybutane
<u>Answer: </u><em>B. Adding more protons to a positively charged body until the number of protons matches the number of electrons</em>
Option B is the appropriate response
<u>Explanation:</u>
Utilising the equivalent number of inverse charges will kill a charged body.
Adding more protons to a decidedly charged body until the number of protons coordinates the quantity of electrons won't kill the body since protons are emphatically charged particles. Adding more protons to an emphatically charged body would make it all the more decidedly charged.
Enabling free electrons to escape from a contrarily charged body will kill since the more negative body leaves the negative electrons.