Answer:
Explanation:
Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:
= 2.25 mol of chlorine
= 0.750 mol of Al.
To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.
So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly 
.
However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be 
.
Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6
130.954g of Fe2O3 in 0.82 mol
Answer:
1) 0.3g Mg
2)0.5g MgO
3)0.2g O
4)0.01mol Mg & 0.01mol O
5)0.01mol MgO
6) Empirical formula MgO
Explanation:
The mass og Mg is obtained by substracting 24.36g from 24.66g:
24.66 - 24.36 = 0.3g Mg
The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.
We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:
*
= 0.2g O
Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO
We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:
*
= 0.01mol O
*
= 0.01mol Mg
The moles of MgO can be obtained from:
*
= 0.01mol MgO
To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.
The result for both number of Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the formula unit of the compound.
The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.
The activity series of metals as well as the electrode potential of metals can be used to compare the reactivity of metals.
<h3>What is used in comparing reactivity of metals?</h3>
The reactivity of metals can be compared using their electrode potentials which is a measures of the ability of the metal to donate electrons to another metal.
When comparing the reactivity of metals, the metal with the lesser negative electrode potential will be more reactive than another with a greater negative or positive electrode potential.
Therefore, the activity series of metals as well as the electrode potential of metals can be used to compare the reactivity of metals.
Learn more about activity series of metals at: brainly.com/question/17469010
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Se - 78
Selenium - 78
or
78/34 Se2-
