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Misha Larkins [42]
3 years ago
12

Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everythi

ng. For example, chlorofluorocarbons, which destroy stratospheric ozone, are not attacked by the hydroxyl radical. Consider the hypothetical reaction by which the hydroxyl radical might react with a chlorofluorocarbon:
OH(g)+CF2Cl2(g)→HOF(g)+CFCl2(g)

Use bond energies to explain why this reaction is improbable. Calculate ΔHrxn of this reaction.

Chemistry
1 answer:
valentina_108 [34]3 years ago
5 0

Find attachments for complete solution

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Which formula represents the compound commonly known as phosphine
Arlecino [84]
I looked up what is the molecular formula for Phosphine and got this: PH3
Hope this helps! Let my know if this was correct.
8 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
Which model of the atom first introduced the idea of a nucleus?
ankoles [38]
The answer is c.) Rutherford model
5 0
3 years ago
Read 2 more answers
If the dehydration reaction of an alcohol is successful. What changes would be seen in the ir spectrum for the product compared
Bess [88]

If the dehydration reaction of an alcohol is successful. The changes would be seen in the IR spectrum for the product compared to the starting material are as,

  • The O-H and  C-O  band is disappear from stating material
  • The addition of a C-C double bond band in the product.

In dehydration reaction of alcohol (  O-H and  C-O bond ) contain , the water molecule ( H_{2}O ) is release from the reactant and C-C double bond is form which is known as alkene in the product .

The reactant and product have different structure. To determine the structure of the compound IR spectroscopy is used. In IR spectrum the peak corresponds to 3400-3600 cm is missing in the product of  dehydration reaction of an alcohol. It means  O-H band is disappear from stating material.

learn about IR SPECTRUM

brainly.com/question/15452269

#SPJ4

4 0
1 year ago
PLEASE HELP ME !!!!!!
Leviafan [203]

Answer: b

Explanation:

By adding heat you are adding more energy

5 0
3 years ago
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