Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
<span>According to octet rule, atoms with an atomic number less than 20 tend to combine with other atom such that both of these atoms have eight electrons in their valence shells, which gives them the same electronic configuration as that of noble gas.
However, there are few compound that donot obey octel rule. Among the elements mentioned above i.e. oxygen and helium obeys octet rule.
In case of nitrogen, oxide of nitrogen (like NO and NO2) have incomplete octet.
While there are few compounds of Br wherein Br has expanded octet. For example, in BrF5, Br has 12 electrons in valence shell. </span>
Answer:
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Answer:
2,4,4-trimethyl-2-pentene yields mixture of 
and 
Explanation:
In ozonolysis (hydrolysis step involve a reducing agent such as Zn, 
etc.), a pi bond is broken to form ketone/aldehyde.
Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.
Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.
Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.
Here, 2,4,4-trimethyl-2-pentene yields mixture of and
Reaction steps are shown below.
