Answer:
Percentage composition of oxygen = 29.69%
Percentage composition of fluorine = 70.37
Explanation:
Given:
Mass of compound = 0.432 grams
Mass of oxygen = 0.128 grams
Find:
Percentage composition of oxygen
Percentage composition of fluorine
Computation:
Percentage composition of oxygen = [0.128/0.432]100
Percentage composition of oxygen = 29.69%
Percentage composition of fluorine = [(0.432 - 0.128)/0.432]100
Percentage composition of fluorine = 70.37
Mg- is the isoelectronic of Na
Answer:
C
Explanation:
Calculate the number of moles of Al2O3 that are produced when 15 mol of Fe is produced
in the following reaction.
C) 45 mol
Answer:
+VE
Explanation:
If we look at the reaction profile pictured in the question, we can easily identify A as the enthalpy of the reaction. The enthalpy of reaction (ΔHrxn) is usually defined as the difference between the total enthalpy (heat content) of the products of a reaction and the total enthalpy (heat content) of the reactants in that reaction.
Looking at the figure, we can see that the enthalpy of products is greater than the enthalpy of reactants, hence ∆Hrxn is positive as stated in the answer above.
Answer:
329.7%
Explanation:
Percent Yield = Actual Yield/ Theoretical Yield x 100%
Percent Yield = 105.5g/32 x 100% = 329.69 ≈ 329.7 %
