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Yakvenalex [24]
3 years ago
15

The amount of water vapor in the air, compared to the maximum amount that can be held at a specific air temperature is called:

Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer: D relative humidity

Explanation: Just took the test.

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A tree trunk floating in the water<br> Balanced<br> Or<br> Unbalanced
mars1129 [50]

Answer:

Balanced

Explanation:

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3 0
3 years ago
What is the activation energy of this reaction? 10 kJ 25 kJ 30 kJ 35 kJ
Nostrana [21]

Answer:

25 kj

Explanation:

8 0
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%

6 0
3 years ago
What is the first thing you notice about a person? .-_-
Ilia_Sergeevich [38]

Answer:

Their smile, I don't know why my eyes just go straight to teeth

7 0
3 years ago
Read 2 more answers
What is the mass ratio of aluminum to oxygen in aluminum oxide, Al2O3?
liubo4ka [24]

Answer:

9 : 8

Explanation:

Aluminum oxide has the following formula Al₂O₃.

Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:

Mass of Al in Al₂O₃ = 2 × 27 = 54 g

Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g

Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:

Mass of Al = 54 g

Mass of O₂ = 48 g

Mass of Al : Mass of O₂ = 54 : 48

Mass of Al : Mass of O₂ = 54 / 48

Mass of Al : Mass of O₂ = 9 / 8

Mass of Al : Mass of O₂ = 9 : 8

Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8

5 0
2 years ago
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