The number of electrons in outer ring for cesium would be ONE 1.
Answer:
39.6 g
Explanation:
The equation of the reaction is;
2Mg(s) + O2(g) --------> 2MgO(s)
To obtain the limiting reactant;
Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles
If 2 moles of Mg yields 2 moles of MgO
1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO
Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles
If 1 mole of O2 yields 2 moles of MgO
0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO
Hence Mg is the limiting reactant.
Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g
Percent yield = 90%
Percent yield = actual yield/theoretical yield * 100
Actual yield = Percent yield * theoretical yield/100
Actual yield = 90 * 44/100
Actual yield = 39.6 g
In terms of the disappearance of the reactants, the rate equations are R= -kd[I-]/dt and -kd[OCl-]/dt.
<h3>What is an ionic reaction?</h3>
The term ionic reaction refers to the reaction that takes place between two ions. In this case, the ionic reaction is; I-(aq) + OCl-(aq) -------> Cl-(aq) + OI-(aq).
The rate equations in terms of the disappearance of the reactants is;
R= -kd[I-]/dt
And
R = -kd[OCl-]/dt
Learn more about ionic reaction:brainly.com/question/12164558
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Answer:
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Explanation:
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