Chemical reaction for acrylic fiber (7th grade chemistry) pllllzzzzzz help ASAP.

Loretta and her friends want to find out which of their pets can run the fastest. Which kind of scientific investigation should

Feliz [49]

A. and D. would be the best pick for this sort of experiment, but maybe (unlikely) B. because you could see how they could react in certain situations, how they react to danger but I suggest A.

Hope this helps you ☁︎☀︎☁︎

7 0

3 years ago

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For any molecule at the surface the resulting surrounding forces equal zero

Katarina [22]

False is the correct answer

4 0

3 years ago

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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:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

How does increasing the temperature affect collisions?

ddd [48]

Combustion Have a great day!

8 0

2 years ago

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Phosphorus can be stable with 12 electrons in its valence structure while nitrogen can never have more than 8 electrons in its v

dolphi86 [110]

Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.

Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.

On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.

8 0

3 years ago