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Tems11 [23]
3 years ago
14

What is the number of valence electrons that most atoms prefer?

Chemistry
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

eight

Explanation:

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What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
Which of the following are indicators of a chemical change? Select all that apply.
Arturiano [62]
<span>Which of the following are indicators of a chemical change? Select all that apply.
</span>
These are the answers:

<span>color change
temperature change
precipitate formation
gas formation

Hope this helps.
</span>
4 0
2 years ago
Read 2 more answers
What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70gmL.
Blizzard [7]

The volume of the piece of aluminum is 1.96 mL

Explanation:

Density is the relationship of the mass of a substance and its volume.

In this case, the mass of aluminum is 5.30 g and the density is 2.70 g/mL

The formula to apply is;

D=M/V where D is density in g/mL, M is mass in g and V is volume in mL

2.70=5.30/V

V=5.30/2.70 =1.96 mL

Learn More

Density of a substance:brainly.com/question/12605423

Keywords: volume, aluminum,density

#LearnwithBrainly

4 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
Which type of compounds are likely to be nonconductors in their solid forms but conductors when they are mixed with water
Blizzard [7]
Ionic compounds. because when in solid form they are held firmly in place therefore they cannot move to conduct electric current.
8 0
3 years ago
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