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Ipatiy [6.2K]
2 years ago
14

Mercury(ii) oxide (hgo) decomposes to form mercury (hg) and oxygen (o2). the balanced chemical equation is shown below. 2hgo rig

ht arrow. 2hg o2 the molar mass of o2 is 32.00 g/mol. how many moles of hgo are needed to produce 250.0 g of o2? 3.906 moles 7.813 moles 15.63 moles 73.87 moles
Chemistry
1 answer:
Verizon [17]2 years ago
3 0

Decomposition reaction results in the formation of the products by splitting the reactants. Moles of mercury(ii) oxide needed are 15.63 moles.

<h3>What are moles?</h3>

Moles are the ratio of mass and the molar mass of the compound or the molecule.

Moles of oxygen are calculated as:

\begin{aligned}\rm n &= \rm \dfrac {mass}{molar\; mass}\\\\&= \dfrac{250}{32}\\\\&= 7.8125 \;\rm moles\end{aligned}

The balanced chemical reaction can be shown as,

\rm 2HgO \rightarrow 2Hg + O_{2}

From the reaction, it can be said that 1 mole of oxygen requires 2 moles of mercury oxide.

Moles of mercury oxide are calculated as:

\begin{aligned}  \rm n HgO& = (7.8125 \;\rm moles \; O_{2}) \times (\dfrac{2 \;\rm moles\;  HgO}{1 \;\rm mole \; O_{2}})\\\\\rm n HgO &= 15.625 \;\rm moles \end{aligned}

Therefore, option C. 15.63 moles of mercury (II) oxide is needed.

Learn more about moles here:

brainly.com/question/26898237

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Serhud [2]
It changes because when the 2 tectonic plates collide they form mountains and volcanoes.

Continental drift causes tectonic plates to either "pull apart" or "crash" into each other.

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8 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
3 years ago
How has the work of chemists advanced cancer treatment?​
jeyben [28]

Answer:

C

Explanation:

it wouldn't be A because that makes no sense

theirs no pill or any type of drug that can slow it down but their is a treatment plan for it

(It could be c or d)

6 0
3 years ago
Read 2 more answers
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AnnyKZ [126]
Answer is: MgF₂, magnesium fluoride.
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6 0
3 years ago
An aluminum block has a density of 2.70 g/mL. If the mass of the block is 24.60 g, find the volume of the substance.
harina [27]

Volume of a substance can be determined by dividing mass of the substance by its density.

That can be mathematical shown as:

Density=Mass/Volume

So, Volume=Mass/Density

Here mass of the substance given as 24.60 g

Whereas density of the substance is 2.70 g/mL

So,

Volume=Mass/Density

=24.6/2.7

=9.1 mL

So volume of the substance is 9.1 mL.

8 0
3 years ago
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