The diatomic gas having a density of 3.164 g/L at STP should be identified. Concept introduction: STP : standard temperature and pressure, it is the condition of 273.2 K and 1 atm . The molar volume of gases at STP condition is similar and it is close to the molar volume of an ideal gas at STP condition. That is, 22.42 L. Hence, at STP the molar volume of a gas is considered as 22.42L. Equation for density is, Density = Mass Volume According to ideal gas equation for molecular mass, Molecular mass = Mass * R * Temperature Volume * pressure
The diatomic gas having a density of 3.164 g/L at STP should be identified. Concept introduction: STP : standard temperature and pressure, it is the condition of 273.2 K and 1 atm . The molar volume of gases at STP condition is similar and it is close to the molar volume of an ideal gas at STP condition. That is, 22.42 L. Hence, at STP the molar volume of a gas is considered as 22.42L. Equation for density is, Density = Mass Volume According to ideal gas equation for molecular mass, Molecular mass = Mass * R * Temperature Volume * pressure
Solution Summary: The author explains that the diatomic gas having a density of 3.164 g/L at STP is Chlorine.
The diatomic gas having a density of 3.164 g/L at STP should be identified.
Concept introduction:
STP: standard temperature and pressure, it is the condition of
273.2K and 1 atm.
The molar volume of gases at STP condition is similar and it is close to the molar volume of an ideal gas at STP condition. That is, 22.42 L. Hence, at STP the molar volume of a gas is considered as 22.42L.
Equation for density is,
Density=MassVolume
According to ideal gas equation for molecular mass,
Michael Reactions
19.52 Draw the products from the following Michael addition reactions.
1.
H&C CH
(a)
i
2. H₂O*
(b)
OEt
(c)
EtO
H₂NEt
(d)
ΕΙΟ
+
1. NaOEt
2. H₂O'
H
H
1. NaOEt
2. H₂O*
Rank the labeled protons (Ha-Hd) in order of increasing acidity, starting with the least acidic.
НОН НЬ
OHd
Онс
Can the target compound at right be efficiently synthesized in good yield from the unsubstituted benzene at left?
?
starting
material
target
If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area.
Be sure you follow the standard ALEKS rules for submitting syntheses.
+ More...
Note for advanced students: you may assume that you are using a large excess of benzene as your starting material.
C
:0
T
Add/Remove step
G
Chapter 5 Solutions
Bundle: Chemistry, 10th + Laboratory Handbook for General Chemistry, 3rd + Student Resource Center Printed Access Card + Student Solutions Manual for ... Access Card for Zumdahl/Zumdahl/DeCoste
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