Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
Answer:
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Explanation:
<u>Step 1:</u> Data given
A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K
Moles of CO2 = 1.27 moles
Moles of CO = 3.04 moles
Moles of Ar = 1.50 moles
<u>Step 2:</u> Calculate total number of moles
Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles
<u>Step 3:</u> Calculate mol fraction Ar
Mol fraction Ar = 1.50 mol/5.81 mol = 0.258
<u>Step 4</u>: Calculate partial pressure
1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Answer:
21.86582KJ
Explanation:
The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.
Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.
<span>100.
ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water
Molarity, M
M = number of moles of solute / volume of solution in liters
number of moles of solute = mass of CHCl3 / molar mass of CHCl3
molar mass of CHCl3 = 119.37 g/mol
number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol
using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters
M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer
Molality, m
m = number of moles of solute / kg of solvent
number of moles of solute = 0.838
kg of solvent = kg of water = 1,000,000 kg
m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer
mole fraction of solute, X solute
X solute = number of moles of solute / number of moles of solution
number of moles of solute = 0.838
number of moles of solution = number of moles of solute + number of moles of solvent
number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles
number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles
X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer
mass percent, %
% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =
% = 10 ^ - 6 % <------- answer
</span>
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.
