FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
Answer:
d. the conjugate base of the weak acid
Explanation:
The strong base (BOH) is completely dissociated in water:
BOH → B⁺ + OH⁻
The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.
By other hand, the weak acid (HA) is only slightly dissociated in water:
HA ⇄ H⁺ + A⁻
The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.
Therefore, the resulting solution will have a pH > 7 (basic).
Ok so First of all we start with the fire. The fire gives off radiation because you can feel the heat through space. The fire also gives of conduction because you put the hotdog on the fire to cook it, and the hotdog will give off steam when it is hot causing it to give of Convection.
There is how cooking a hotdog over a fire uses all three heat transfer
Answer and Explanation:
As the temperature of the substances increases, the average energy of the molecules increases, and average energy of attraction between the molecules decreases consequently intermolecular spacing between the molecules increases. As a result, a substance change in succession from gas to liquid to solid.
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
Hence, the correct answer is Option D.
