Answer:
A. Semi Permeable.
Explanation:
Cell membranes are semi permeable.
Answer: 4-allylanisole
Explanation: The doublets behind the 7 ppm belongs to the
para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group
A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.
Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.
The Predicted Number is attached from a chemical database along with their peaks information
Standard temperature and pressure (STP) means a temperature of 0°c and a pressure of 1 atmosphere (atm). The molar gas volume is used to convert between the number of moles of a gas and the volume of the gas at STP. One mole of a gas occupies a volume of 22400 cm³ or 22.4 liters at STP according to the molar gas volume.
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
