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morpeh [17]
3 years ago
9

Chemical reactions constantly take place all around us—and even inside us. Many

Chemistry
1 answer:
Alika [10]3 years ago
5 0

Answer:

I don't think you can go on without doing a reaction since you touch really anything so no you can not go without any chemical reaction I think this is the answer

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The equilibrium constant, Kp, for the following reaction is 0.636 at 600K.
natulia [17]

Answer: The equilibrium partial pressure of Cl_{2} is 0.964 atm.

Explanation:

Given: K_{p} = 0.636

P_{COCl_{2}} = 0.836 atm

P_{CO} = 0.551 atm

The given reaction equation is as follows.

COCl_{2}(g) \rightleftharpoons CO(g) + Cl_{2}(g)

Formula used to calculate the partial pressure of Cl_{2} is as follows.

K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}

Substitute the values into above formula as follows.

K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}\\0.636 = \frac{0.551 atm \times P_{Cl_{2}}}{0.836 atm}\\P_{Cl_{2}} = 0.964 atm

Thus, we can conclude that the equilibrium partial pressure of Cl_{2} is 0.964 atm.

3 0
3 years ago
Propose a mechanism for the following reaction.
raketka [301]
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation. 
5 0
3 years ago
Can someone help me on some of these question please : )
frez [133]
X-Ray used to Take images of teeth and bones.
Ultraviolet night vision goggle
4 0
4 years ago
3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on
AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of  polysaccharide  and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis;  starch should give a positve test, while Diasaccharides should give negative rest.

3 0
3 years ago
What are the phase change processes taking place when a substance is in equilibrium between liquid and gas phases? A. Melting an
lawyer [7]
C. <span>Evaporating and condensing. When the liquid becomes gas is called evaporation. When gas become liquid it's condensation. </span>
3 0
3 years ago
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