"<span>A. The polar regions have long winters and short, cool summers." would be the correct answer.
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Answer:
It will take 0.0205s for 79% of cyclobutane to decompose
Explanation:
Step 1: Data given
Temperature = 1000 °C
Reaction is first order
rate constant = 76*1/s
The initial cyclobutane concentration is 1.63M
How long (in seconds) will it take for 79% of the cyclobutane to decompose?
When 79% decomposes, there will remain 21 %
ln ([A]0 / [A]t) * = k*t
There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.
⇒ with [A]0 = the original amount = 100 % =1
⇒ with [A]t = the amount after it's decomposed = 21 % =0.21
⇒ k = the rate constant = 76/s
⇒ t= the time needed = ?
ln (100/21) = 76t
t = 0.0205 s
It will take 0.0205s for 79% of cyclobutane to decompose
Answer:
56.17% is percent composition of Sb in the molecule
Explanation:
Percent composition is the percent in mass of each element present in a particular molecule.
In SbF₅, there is 1 mole of Antimony -Molar mass: 121.76g- per 5 moles of fluorine -Molar mass: 19g/mol-. In a basis of 1 mole, the mass of Sb and F is:
<em>Mass Sb:</em>
1mol * (121.76g/mol) = 121.76g
<em>Mass F:</em>
1 mol SbF₅ = 5 moles F * (19g / mol) = 95g
<em>Total mass:</em>
121.76g + 95g = 216.76g
And percent composition of Sb:
121.76g / 216.76g * 100 =
<h3>56.17% is percent composition of Sb in the molecule</h3>
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