the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
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FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
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In the reaction of magnesium metal with hydrochloric acid, we can determine that the magnesium metal has reacted completely when the magnesium metal is gone.
<em>In the reaction of magnesium metal with hydrochloric acid, how do you determine when the magnesium metal has reacted completely? select all that apply.</em>
- <em>All of the water has evaporated. </em>
- <em>The dilute HCl is gone. </em>
- <em>The magnesium metal is gone. </em>
- <em>Gas bubbles are no longer produced.</em>
Magnesium reacts with hydrochloric acid according to the following equation.
Mg(s) + 2 HCl(aq) ⇒ MgCl₂(aq) + H₂(g)
How can we determine when the magnesium metal has reacted completely?
All of the water has evaporated. NO. Water is not part of the reaction.
The dilute HCl is gone. NO. It indicates that HCl reacted completely.
The magnesium metal is gone. YES. If Mg is gone means it reacted completely.
Gas bubbles are no longer produced. NO. The bubbles, produced by hydrogen, may not be produced due to the insufficiency of HCl.
In the reaction of magnesium metal with hydrochloric acid, we can determine that the magnesium metal has reacted completely when the magnesium metal is gone.
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Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.Ca²⁺ is formed as,
Ca → Ca²⁺ + 2 e⁻
These two electrons are accepte by Sulfur as,
S + 2 e⁻ → S²⁻
So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs.
Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.
Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.
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I think the most appropriate answer is B !
as electrolysis is non spontaneous ! and it is also a redox rkn !