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Sergio [31]
3 years ago
9

Have people considered that aliens could possibly live without oxygen and water and we are looking in the wrong direction for a

new life form?
This is astronomy, not chemistry. Sorry, they do not have the option of astronomy.
Chemistry
1 answer:
bija089 [108]3 years ago
8 0
I have never heard anyone ever say anything about aliens only living by using the same metabolistic processes as humans... But.. there may likely be another planet out there that does use oxygen, although at a different percentage from Earths. Anyways, even if we do find "other" life forms, who's to say that they're on the same level as us? Have you ever considered the fact that ants to us are like we are to other life forms out there and maybe they don't want or can't communicate with us because they see themselves as superior?

Hope that helps!
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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with
Pavlova-9 [17]

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a <u>Grignard reagent</u> (ethylmagnesium bromide), therefore we will have the production of a <u>carbanion</u> (step 1). Then this carbanion can <u>attack the least substituted carbon</u> in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the <u>treatment with aqueous acid</u>, when we add acid the <u>hydronium ion</u> (H^+)  would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be <u>attacked by the negative charge</u> produced in the second step to produce the final molecule: <u>"Pentan-2-ol".</u>

See figure 1

I hope it helps!

5 0
3 years ago
What causes metal to emit light?​
Flauer [41]
Heating an atom excites its electrons and they jump to higher energy levels. When the electrons return to lower energy levels, they emit energy in the form of light. ... Every element has a different number of electrons and a different set of energy levels. Thus, each element emits its own set of colours.
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_H3PO4 + _HCl ---&gt;_PCl5 + _H2O
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Which of the following are excited in ultraviolet photoelectron spectroscopy?
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D. Valance Electrons

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