D. Two electrons in its first energy level; eight electrons in its second energy level; six valence electrons in its outermost energy level.
Please correct me if I'm wrong!! :)
<u>Answer:</u> The limiting reagent in the reaction is bromine.
<u>Explanation:</u>
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
Given values:
Moles of iron = 10.0 moles
Moles of bromine = 12.0 moles
The chemical equation for the reaction of iron and bromine follows:
By the stoichiometry of the reaction:
If 3 moles of bromine reacts with 2 moles of iron
So, 12.0 moles of bromine will react with = 
of iron
As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Hence, bromine is considered a limiting reagent because it limits the formation of the product.
Thus, the limiting reagent in the reaction is bromine.
Answer:
4.56 X 10^ -4 g/mL
Explanation:
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.
(7.6 X10^-4 gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g
this is dissolved )in 10 m L=45.6 X 10^-4 g/ 10
4.56 X 10^ -4 g/mL
check
6/10 =0.6
4.56/7.6 = o.,6
Answer:
0.719M AgNO₃
Explanation:
Based on the reaction:
MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂
<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>
<em />
To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:
<em>Moles AgNO₃:</em>
<em />
Moles of MgBr₂ are:
50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.
As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:
0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =
0.0100 moles of AgNO₃ are in the solution.
And molarity is:
0.0100 moles AgNO₃ / 0.0139L =
<h3>0.719M AgNO₃</h3>
