However <em>trans</em>-2-Butene does not give a characteristic peak in 1620-1680 cm⁻¹ region but still the presence of carbon double bond carbon can be detected by detecting following peaks in IR Spectrum.
1) 3010-3100 cm⁻¹:
As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.
2) 675-1000 cm⁻¹:
Another peak which is given by the bending of =C-H (C-H) bond with strong intensity will appear in the range of 675-1000 cm⁻¹.
Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
Air is a mixture of several gasses
I think it say go rggdggy and do something with one of those other things and get a work order and get the t off and u I know I have
<h3>
Answer: A) 3.5 mol/L</h3>
Explanation:
To determine the molarity, we have to find the number of moles in the volume given, and then extrapolate to find the number of moles that would be in 1 L.
<u>Determine the moles in the given volume</u>
moles of LiCl = mass ÷ molar mass
= 139.9 g ÷ 42.39 g/mol
= 3.30 mol
<u>Find the moles in 1 L</u>
Since 930 mL of LiCl = 3.30 mol
then 1000 mL of LiCl = (3.30 mol × 1000 mL/L) ÷ 930 mL
= 3.55 mol/L
