Answer:
domain bacteria
Explanation:
Salmonella and E. coli are same in the sense that they are both bacteria,
Part a)
Balanced equation:
2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)
Part b)
ΔH for the reaction at 25°C
ΔH = Enthalpy change of products - Enthalpy change of reactants
= [(2 mol) * (-393.509 kJ/mol) + (4 mol)*(-241.83 kJ/mol)] - [(2 mol) * (238.4 kJ/mol) + (3 mol) *(0 kJ/mol)] = - 1277.5 kJ
Part c)
ΔS for the reaction at 25°C
ΔS = entropy change in products - entropy change in reactants
= [(2 x 213.7) + (4 x 188.8)] - [(2 x 127.2) + (3 x 205.1)] = 312.9 J K⁻¹
Part d)
ΔG for the reaction at 25°C
ΔG = ΔH - (T * ΔS)
= - 1277.5 x 10³ J - (298 x 312.9 J K⁻¹) = -1370.7 kJ
Answer:
See answer below
Explanation:
In this case, let's draw the butane molecule:
CH₃ - CH₂ - CH₂ - CH₃
According to what the exercise states, we removed an atom of hydrogen from the frist carbon. This could be any of the terminals. I'll grab the first from left to right.
CH₂⁺ - CH₂ - CH₂ - CH₃
When this happens, the atom of carbon is lacking one space and it forms a carbocation.
Followed this step, an hydroxile group replace the atom of hydrogen. The hydroxile is the OH, and when we have an alkane with an OH group in the molecule, we are actually converting this molecule into an alcohol, therefore the molecule formed is:
<h2>
OH - CH₂ - CH₂ - CH₂ - CH₃</h2><h2 />
Hope this helps
Many electrophilic aromatic halogenations require the presence of an aluminum trihalide as a catalyst. We generally acetylated the amino group as protection. Now, this acetanilide can be brominated at Ortho or para position. An atom that is attached to an aromatic system usually hydrogen is replaced by an electrophile is an organic reaction which is called Electrophilic aromatic substitution. There are what you called important electrophilic aromatic substitutions they are aromatic nitration, aromatic sulfonation, aromatic halogenation and acylation and alkylating Friedel-Crafts reaction. Aromatic bromination is an electrophilic aromatic substitution (EAS) reaction, which will require benzene to act as a nucleophile to acquire an electrophile. Therefore, any directing groups that activate the ring will make it react more quickly with respect to aromatic bromination. Acetanilide is a moderately-activated ring <span>having a decent EWG.</span>
I think thee correct answer from the choices listed above is option D. <span>When a physical change in a sample occurs, composition of the sample does not change. It stays the same. Also, the properties of the sample will still be the same. Hope this answers the question.</span>
