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abruzzese [7]
3 years ago
8

You are conducting a weathering study and your solution analysis indicates that you have 1.8 X 10-3 M silica and a pH of 3. You

plug this data into your chemical speciation computer program and indicates that quartz precipitates. Do you think this will happen
Chemistry
1 answer:
Basile [38]3 years ago
3 0

Answer:

I do not think it will happen

Explanation:

The result is not general and can only occur when there is higher temperatures and pressure. Only when there's high temperature and pressure, will the solution precipitate.

More so; kinematically, the result is limited though it can happen in thermodynamics environment.

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Static Friction, Sliding Friction, Rolling Friction, and finally Fluid Friction.
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The molecule NH3 contains all single bonds.<br><br> true <br> false
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<h3><u>Answer;</u></h3>

True

<h3><u>Explanation</u>;</h3>
  • The molecule NH3 contains all single bonds.
  • NH3 has a three single covalent  bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
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Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

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