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3 years ago

Please help!!

Chemistry

1 answer:

sammy [17]3 years ago

3 0

Answer:

Explanation:

because since the outer shell has 5 electron then add 5 and 10 to get 15

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For the following exothermic reaction, A+B⇌C+D + Heat, if the temperature is increased, the reaction will experience a shift to

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(Left) I’m pretty sure

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3 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

Is o-toluic acid soluble in ether, NaOH?, Is Napthalene soluble in NaOH?

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Naphthalene is insoluble in NaOH.

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3 years ago

A sample of the compound magnesium oxide is synthesized as follows. 60. g of magnesium is burned and produces 100. g of magnesiu

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Answer:

30 g of magnesium would be combined with 20 g of oxygen. The law used solving this problem is the Lavoisier Law of conservation of mass.

Explanation:

If 60 g of magnesium combines with 40 g of oxygen to make 100 g of magnesium oxide, then 30 g of magnesium will combine with 20 g of oxygen to make 50 g of magnesium oxide.

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The following equation represents the proportions of the substances:

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