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Kamila [148]
3 years ago
13

Question 14 (4 points)

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
7 0

Answer:

Sodium Phosphide + Fluorite --------- Sodium Fluoride + Calcium Phosphide

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Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy
SVETLANKA909090 [29]

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ


2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm


3. Repeat the calculation in 1, but using n=5.


4. Repeat the calculation in 2 using the energy calculated in 3.

7 0
3 years ago
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Calculate the average bond order for a p−o bond (such as the one shown in blue) in a phosphate ion. express your answer numerica
PtichkaEL [24]
<span>Answer: 1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>
5 0
3 years ago
Which of the following phase changes describes what is
stich3 [128]

Answer:

I think the answer is boiling

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
Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)​
Ket [755]

Answer:

Na2SO4 is equal to 0.0070401642780093 mole.

Explanation:

5 0
3 years ago
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