[A]0= Initial concentration
t1/2= half life
[A]= final concentration
k= rate constant
Answer:
54.30 grams acetone
Explanation:
(0. 935 mol)*(58.08 g/mole) = 54.30 grams acetone
the correct IUPAC name of the compound is 1-Butanal.
<h3>What are IUPAC names?</h3>
It is a system of naming organic compounds based on the longest carbon-to-carbon single bonds. It does not matter whether these longest chains are continuous or in a ring.
Thus, when the compound with the chemical formula, CH3-CH2-CH2CHO is considered. The longest carbon-to-carbon chain is 4. The 1st carbon carries a functional group known as an aldehyde.
Aldehydes are equipped with the carbonyl group and have the general formula R−CH=O. They are also sometimes referred to as formyl.
Aldehydes are named after their parent alkane chains with a slight modification. The 'e' is replaced with 'al'
The aldehyde in this case has four carbons. This means that the parent alkane is Butane. Therefore, the name of the compound will be 1-Butanal.
More on IUPAC names can be found here: brainly.com/question/16631447
#SPJ1
Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm
