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andrey2020 [161]
2 years ago
8

If 32 L of hydrogen react with excess oxygen at STP to form water, how much energy will be released?

Chemistry
1 answer:
Fudgin [204]2 years ago
4 0

Answer:

??????????????????

Explanation:

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Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart
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Answer:

B. Valence Electron Pairs

Explanation:

Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.

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A gas at 400 k is placed in a 4.0 l moveable piston. The gas is cooled to 252 k. What is the new volume of the piston?
katrin [286]

Answer:

2.52L

       

Explanation:

Given parameters:

T₁  = 400K

V₁  = 4L

T₂ = 252K

unknown

V₂ = ?

Solution:

To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.

Mathematically,

               \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

Substitute and solve for V₂

           \frac{4}{400}  = \frac{V_{2} }{252}

        V₂ = 2.52L

         

6 0
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When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

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.

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