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Naddika [18.5K]
3 years ago
9

Heterogeneous mixture are often separated by

Chemistry
1 answer:
algol [13]3 years ago
4 0
The answer is Filtration

Hope this helps
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100

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An object has a negative charge if its atoms what
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An object has a negative charge if it's atoms have more electrons. On the other hand if an object had more protons, then it would have a positive charge.

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6 0
3 years ago
Write a balanced equation to show the reaction of gaseous ethane with gaseous oxygen to form carbon monoxide gas and water vapor
Korolek [52]
<span>2 C2H6(g) + 5 O2(g) --------> 4 CO(g) + 6 H2O(g)

</span>
7 0
4 years ago
1. What is the best method for removing water from a hydrated compound?
Helga [31]
The traditional method is to heat the compound in an oven, on a hot plate or over a Bunsen burner. Waters of hydration are loosely bound in the compound and can be driven off at temperatures below the melting point of the compound. So the answer is heating.
6 0
3 years ago
This question involves two calculations. The answer to the first part will be
ozzi

Answer :

(a) The mass of Al_2O_3 produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of Al = 85.1 g

Molar mass of Al = 27 g/mol

First we have to calculate the moles of Al

\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol

Now we have to calculate the moles of Al_2O_3

The balanced chemical equation is:

4Al+3O_2\rightarrow 2Al_2O_3

From the reaction, we conclude that

As, 4 moles of Al react to give 2 moles of Al_2O_3

So, 3.15 moles of Al react to give \frac{2}{4}\times 3.15=1.58 mole of Al_2O_3

Now we have to calculate the mass of Al_2O_3

\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3

Molar mass of Al_2O_3 = 102 g/mole

\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g

Therefore, the mass of Al_2O_3 produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%

Therefore, the percent yield of the reaction is, 72.5 %

8 0
3 years ago
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