Lord Kelvin, were he alive today, would be considered a Thermochemist. Thermochemistry is interested in the role of heat in chemical reactions. This includes the role of heat both as a biproduct of chemical reactions and a facilitator.
Kelvin's description of absolute zero is an important concept in thermochemistry. At absolute zero, there is no movement of molecules, and no energy available facilitate chemical reactions.
It’s b for sure that’s the answer
Answer:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Explanation:
We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.
HNO₃ is a strong acid and will have the largest hydronium concentration.
HCN Ka = 6.2 x 10⁻¹⁰
HNO₂ Ka = 4.0 x 10⁻⁴
HClO Ka = 3.0 x 10⁻⁸
The ranking from smallest to largest hydronium concentration will then be:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
Answer:
pH 4 represents an acidic