Answer:
Explanation:
C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"?
Based on n+1 rule. Here n=2 (identical beta protons).
2+1=3
So the multiplicity of alpha proton is triplet, .
D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal?
The meaning of the adjective is this: the multiplicity of beta protons is singlet only (no spliting) in absence of alpha proton . But beta protons splits as doublet (n=1) in the presence of alpha proton,
E) How many bonds connect these "splitting next door neighbors"?
There are 3 bonds in between alpha and beta protons in a molecule.
F) What is the multiplicity of the Proton-betas' signal?
Following the n+1 rule, here n=1 (1 alpha proton) so 1+1=2. Hence it is a doublet.
Answer:
It is the minimum amount of analyte that produces a signal which can be measured with reasonable accuracy - LOQ
The concentration is equal to three times the standard deviation of the signal from the blank divided by the slope of the calibration curve - LOD
The concentration is equal to 10 times the standard deviation of the signal from the blank divided by the slope of the calibration curve - LOQ
It is the minimum amount of analyte that produces a signal that is significantly different from the blank - LOD
Explanation:
We define the limit of detection has the lowest amount of analyte that produces a signal that is significantly different from a blank solution ( the absence of the substance). It is calculated as three times the standard deviation of the signal from the blank divided by the slope of the calibration curve.
The limit of quantitation (LOQ) is defined as the minimum amount of analyte that produces a signal which can be measured with reasonable accuracy. It is measured as 10 times the standard deviation of the signal from the blank divided by the slope of the calibration curve.
-1 charge = the atom has gained one electron
9+1=10
Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
