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3 years ago

PLEASE HELP AND DONT YOU DARE GIVE ME THE LINK

Chemistry

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer:

animals and other living things

Explanation:

All living organisms have been classified and grouped into a hierarchical system of classification. Kingdom is the highest level in this system comprising of 6 other levels. However, there are different kingdoms that harbors different organisms. The kingdoms are: animalia, plantae, protista, monera, fungi etc.

However, among other kingdom of organisms, organisms in the Kingdom Animalia are the only ones that possesses muscle tissues etc. This shows a contrast between animals and other living things.

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Explanation:

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3 years ago

2. When vinegar is added to eggshells, carbon dioxide gas is produced.Why?

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3 years ago

ethanol is a common laboratory solvent and has a density of 0.789 g/mL. what is the mass, in grams, of 151 mL of ethanol

natta225 [31]

V = 151 mL = 151 cm³
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--------------------------------------

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3 0

3 years ago

Read 2 more answers

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

3 years ago