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3 years ago

What does the Atomic Number tell you? the amounr of electronstprorons

Chemistry

1 answer:

tankabanditka [31]3 years ago

6 0

Answer

Atomic number or proton number tells us the number of protons found in the nucleus of every atom of that element

Explanation:

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The molar mass is 294.1527 g/mol

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3 years ago

How much work is required to lift a 0.500 kg block 18.5 m?

kipiarov [429]

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

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3 years ago

What can you extract from a refinery

Dmitrij [34]

Answer:

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Explanation:

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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

Where is the atmosephere the most dence

likoan [24]

The Earths Surface, it’s also known as the troposphere also dense* lol

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3 years ago

Read 2 more answers