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cupoosta [38]
2 years ago
11

Plz answer asap The number of moles in 5.82 x 10^22 sulfate ions (SO4^2-)

Chemistry
1 answer:
Elden [556K]2 years ago
3 0

Answer:

i don't know sorry

Explanation:

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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
Why cant (NaCl) light the bulb until the NaCl is dissolved in water. *
Aneli [31]

Answer:

NaCl will only conduct electricity in solutions

Explanation:

For electrical conduction, free mobile electrons as seen in most metals must be present or ions which are charged particles must be available for solutions and molten substances.

  • Sodium chloride is an ionic compound without free mobile electrons or ions despite being ionic.
  • It will maintain a subtle and unique charge stability when in solid form.
  • In solid, the ions are not free to move and remain locked up in the solid mass.
  • When introduced into a solution, the ions becomes free to move and this will aid electrical conduction.
6 0
3 years ago
In the boxes below, draw three particle diagrams representing the sample
Alexxandr [17]

Answer: 2

Explanation: bencause they are changing

5 0
3 years ago
How many millimeters of 5 M NaCl are required to prepare 1500 mL of 0.002 M NaCl?
timurjin [86]

Answer:

The correct answer is 0.6 mL.

Explanation:

We use the mathematical expression:

Ci x Vi = Cf x Vf

Where Ci is the initial concentration (5 M); Cf and Vf refers to final concentration (0.002 M) and final volume (1500 mL). With the given data, we calculate the initial volume (Vi):

Vi = (Cf x Vf)/Ci = (0.002 M x 1500 mL)/(5 M) = 0.6 mL

Therefore, we need 0.6 mL of 5 M NaCl to prepare the solution with the requested dilution.

3 0
3 years ago
The flame from the stove the thermal energy
MatroZZZ [7]
The temp is the flam
7 0
3 years ago
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