Answer:
The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.
Explanation:
The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.
TWA = (125x2+88x2+112x2.5+70x1.5)/8.
The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.
Solution :
"Aldol" stands for the abbreviation, aldehyde and alcohol. When a ketone or an aldehyde's enolate reacts with the carbonyl of a molecule at the alpha carbon, under the acidic or basic conditions so as to obtained the ketone or β-hydroxy aldehyde, is known as an aldol reaction.
For the conversion of the aldol addition product of a 3-hydroxy-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propanone to an aldol condensation product of (E)-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propenone, the mechanism is given in the diagram a below :
Answer:
n = 2.1 mol
Explanation:
Given data:
Number of moles of gas = ?
Volume of gas = 56.3 L
Pressure of gas = 0.899 atm
Temperature of gas = 20°C (20+273 = 293 k)
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
0.899 atm × 56.3 L = n × 0.0821 atm.L/ mol.K × 293 k
50.614 atm.L = n × 24.055 atm.L/ mol
n = 50.614 atm.L / 24.055 atm.L/ mol
n = 2.1 mol
Metal
On the periodic table its atomic number is 89
The Semi-metals aka the metalloids are atomic umbers 5,14,32,33,51,52,and 85
The Metals are everything that is located to the left of the metalloids excluding Hydrogen(H) and Helium (He)
Everything to the right of the metalloids are non-metals including Hydrogen (H) and Helium (He)
Answer:
[AB] = 0.66M
Explanation:
rate = change in concentration/time
Initial concentration of AB = 1.5M
Let the concentration of AB after 10.3s be y
Therefore, rate = 1.5 - y/10.3
The rate equation is given as
rate = k[AB]^2 = 0.2y^2
0.2y^2 = 1.5 - y/10.3
2.06y^2 = 1.5 - y
2.06y^2 + y - 1.5 = 0
Using the quadratic formula
y = [-1 + or - √(1^2 -4×2.06×-1.5)/2(2.06)]
The value of y must be positive
y = (-1 + √13.36)/4.06 = -1+3.66/4.06 = 2.66/4.06 = 0.66
Concentration of AB after 10.3s is 0.66M
