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Leviafan [203]
3 years ago
5

Can someone help me lol;) WILL MARK BRAINLIEST

Chemistry
1 answer:
inn [45]3 years ago
4 0

Answer:

See explanation

Explanation:

1 ) Al has physical and chemical properties, similar to Boron, because It's also part of group 13, also called the boron group. The elements are characterized by having three valence electrons.

2) Lithium is in group 1. This means it has 1 valence electron.

Nitrogen has 7 electrons. The first shell can only have 2 electrons, so the second and outer shell has 5 electrons. Those 5 electrons are the 5 valence electrons;

Neon is a noble gas, it's in the stable noble gas configuration. This means it has 8 valence electrons.

Oxygen = non-metal because it's a gas

Silicon = metalloid

Helium = non-metal because it's a gas

Arsenic = metalloid

Sodium = it's an (alkali) metal, It is physically silver colored and is a soft metal of low density. Pure sodium is not found naturally on earth because it is a highly reactive metal

Chlorine = non-metal

Tin = Is a soft,  ductile and highly crystalline silvery-white metal

Boron = it does conduct electricity. Its chemical properties are mostly, but not entirely, and non-metallic

lead = is a heavy metal that is denser than most common materials

Radon = metalloid

Cesium = It is a soft, silvery-golden alkali metal

Carbon = non metal

Period

⇒ Horizontal from left to right

⇒ same number of atomic orbitals

⇒period = row

Group

⇒ vertical colums

⇒same number of valence electrons

⇒ same physical and chemical properties

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Explanation:

M=D times V

Answer-3,633.84g

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What is the molecular formula for a monocyclic hydrocarbon with 14 carbons and 2 triple bonds?
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The molecular formula for a monocyclic hydrocarbon with 14 carbons and 2 triple bond is C₁₄H₂₀

<h3>Molecular formula</h3>

A formula that gives the number of atom of each element present in a one molecule or a compound.

<h3>Monocyclic hydrocarbons</h3>

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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
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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