On the periodic table, the size of an element's ionic radius follows a predictable trend. The ionic radius increases as you shift down a column or community. This is because every row introduces a new shell of electrons. The ionic radius decreases over a row or time, going from left to right.
<em><u>Ionic bonds</u></em> <span>are the type of bonds where there is </span><u>transfer</u><span><u> </u>of electrons from one atom to another. The electrons are removed and from one atom and attached to another. A good example is salt which is composed of sodium and chlorine. Sodium readily loses one of its electrons and chlorine readily accepts it. Before losing the electron, sodium has a positive charge, but then becomes negatively charged after giving up the electron. Chlorine has a positive charge before gaining the electron but becomes negatively charged after gaining the electron. These opposite charges between sodium and chlorine attract the two elements together to form the ionic bond.</span>
Answer:
The answer to your question is limiting reactant = Fe
Explanation:
Data
moles of Fe = 2
moles of O₂ = 6
Process
1.- Write the balanced chemical reaction
Fe + O₂ ⇒ Fe₂O₃
Reactants Elements Products
1 Fe 2
2 O 3
4Fe + 3O₂ ⇒ 2Fe₂O₃
Reactants Elements Products
4 Fe 4
6 O 6
2.- Use the balanced chemical reaction to determine the limiting reactant.
Theoretical proportion of reactants 4 moles Fe / 3 moles O₂ = 1.33
Experimental proportion of reactants = 2 moles Fe/6 moles O₂ = 0.333
3.- Conclusion
As the experimental proportion was lower than the theoretical we conclude that the amount of Fe was lower in the experiment which is the limiting reactant.
If 162.35 g aluminum hydroxide are dissolved in 6750 mL of solution, the concentration of the solution is 0.308M
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated using the following formula:
Molarity = no of moles ÷ volume (L)
However, the number of moles in 162.35g of Al(OH)3 must be calculated as follows:
Molar mass of Al(OH)3 = 78g/mol
no of moles = 162.35g ÷ 78mol
no of moles = 2.08mol
Molarity = 2.08mol ÷ 6.75L
Molarity = 0.308M
Therefore, if 162.35 g aluminum hydroxide are dissolved in 6750 mL of solution, the concentration of the solution is 0.308M.
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