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2 years ago

Sediment is a rock material that forms where rocks and broken down into smaller pieces or ____ in water as rocks erode.

Chemistry

1 answer:

klio [65]2 years ago

3 0

Answer:

Dissolves

Explanation:

Sediment is a rock material that forms where rocks are broken down into smaller pieces or dissolved in water as rocks erode.

They are products of rock weathering.

Weathering can be defined as the physical disintegration and chemical decomposition of rocks to form sediments and soil.
The first part of the statement describes physical disintegration of rocks where they get broken down into smaller chunks.
The other part simply denotes chemical weathering of rocks which lead to solution of some minerals in the rock.
Both forms are two ways by which rock weathering takes place.

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Carbon dioxide (CO2)
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3 years ago

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2 years ago

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

guajiro [1.7K]

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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3 years ago

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Answer:

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3 years ago

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3 years ago