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3 years ago

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In Mexico City, what are the two things that are causing most of the pollution? If you answer with an explanation I will give yo

u Brainlist PLEASE HELP ITS DUE IN 30minutes

1 answer:

Andre45 [30]3 years ago

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Tornados and fires burning down houses

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What is the lewis structure for CH2ClCOO-

FrozenT [24]

Answer: The Lewis structure of Chloroacetate can be found at the attachment below.

Explanation:

CH2ClCOO- The chemical compound is called Chloroacetate.

Reference link for the Chloroacetate structure.

https://www.google.com/search?q=lewis+structure+for+CH2ClCOO-&prmd=ivn&sxsrf=ALeKk03mQcLiY-q5pEriMR0_26ZTXLjmJg:1589680325594&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjfxPPY5LnpAhVloXEKHeAwD-wQ_AUoAXoECA4QAQ&cshid=1589680746615&biw=360&bih=559&dpr=3#

5 0

3 years ago

When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.

Lubov Fominskaja [6]

Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

3 0

3 years ago

What is the sum of protons and neutrons in an atomic nucleus?

Archy [21]

It is the atomic number. So it is D.

6 0

3 years ago

Consider X + Z -> ?

Licemer1 [7]

Answer:

C

Explanation:

Element Z acquires two electrons from element X for stability

4 0

3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

<u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

<u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

<u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

<u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

<u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

4 years ago