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3 years ago

What is an astronomical unit? Why is this unit used to measure distances in the solar system?

2 answers:

7 0

An astronomical unit is a unit of measurement equal to 149.6 kilometers, the mean distance from the center of the earth to the center of the sun.

1 astronomical unit = 92955807.3 miles

VashaNatasha [74]3 years ago

3 0

A astronomical unit is to measure light years.

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What is the use of the crystallisation process?

ioda

to obtain pure salt from seawater.
in order to obtain pure crystals of a substance from an impure mixture.
obtain pure alum crystals from an impure alum.

5 0

2 years ago

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A company employs several high school students in its food-packaging factory. All the students are below 17 years of age. Which

leonid [27]

Answer:

Fair Labor Standards Act

Explanation:

It is correct. I just took the test on Clever

8 0

2 years ago

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**CHEMISTRY!! PLEASE HELP ASAP!!**

Minchanka [31]

Calcium = 1
Sulfur = 1
Oxygen = 4

6 0

3 years ago

Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these

deff fn [24]

The formation of nitric acid from nitrogen, hydrogen, and oxygen can be written as,

N₂ + H₂ + 3O₂ --> 2HNO₃

The net enthalpy of formation of nitric acid is calculated by,
Hrxn = Hproduct - Hreactant

Since all the reactants are in their elemental forms, the simplified equation would be,

Hrxn = Hproduct

Substituting,

Hrxn = (-186.81 kJ/mol)(2 mols)

<em>Answer: -372.42 kJ</em>

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