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2 years ago

HH or Hh or hh??? help asap

Chemistry

2 answers:

enot [183]2 years ago

7 0

The answer is genotype because those are the letters .

Vikki [24]2 years ago

5 0

Answer:

HH BUT IAM NOT SURE

Explanation:

hope it helps

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Hard water contains relatively large concentrations of

ycow [4]

Answer:

D.

All of these ions

Explanation:

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3 years ago

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4 AlF3 + 3 O2 ----------> 2Al2O3 + 6 F2

anyanavicka [17]

Explanation:

6 F2------->4 AlF3

F2-----------> 4/6 AlF3

8.25 F2 ---------> 4×8.25/6 AlF3

so 5.5 moles

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3 years ago

What molecules can be made using hydrogen, carbon, and nitrogen?

Yuki888 [10]

Answer:

Ammonia

Explanation:

Ammonia is the simplest possible molecule made with nitrogen and hydrogen. Methane is the simplest possible molecule made of carbon and hydrogen. Methanol is like methane, but it also has one oxygen atom as well.

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2 years ago

When an atom of oxygen (O) forms an anion by gaining two electrons, what is the anion’s charge?

Rudiy27

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7 0

2 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