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3 years ago

Which of the following describes all of the types of water found in the hydrosphere? ( A) all frozen water on the Earth

Chemistry

2 answers:

larisa86 [58]3 years ago

4 0

Answer:

all of the water on the Earth

Explanation:

ddd [48]3 years ago

4 0

Answer:

(C)all of the water on the Earth

Explanation:

The hydrosphere consists of all the water found on earth. The hydrosphere include water found in the cloud, fresh water bodies and salt water bodies.

The hydrosphere is interlinked via the water cycle. Water leaves the earth surface to the cloud, condenses and falls again to the earth the ground.

The water cycle shows the free flow of water within the hydrosphere.

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3 years ago

Write the coefficients for the reactant and product.

balu736 [363]

<h3>Answer:</h3>

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

<h3>Explanation:</h3>

The equation for the reaction shown represents the combustion of ethane.

C₂H₆ + O₂ → CO₂ + H₂O

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4 0

3 years ago

Read 2 more answers

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Nutka1998 [239]

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3 years ago

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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