Answer:
Please see the complete formt of the question below
Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.
MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)
According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.
The answer to the above question is
The limiting reactant is the MnO₂
Explanation:
To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂
Molar mass of MnO₂ = 86.9368 g/mol
Molar mass of HCl = 36.46 g/mol
From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂
However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed
The mass of ni(NO3)2 that dissolved in 25.0 ml of 0.100m ni(NO3)2 solution is calculated as follows
fin the number of moles = molarity x volume in liters
=25 x0.100/ 1000= 2.5 x10^-3 moles
mass = mass x molar mass
= 2.5 x10^-3 moles x 182.71 g/mol = 0.457 grams
Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
Answer:
At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8
Explanation:
HBCG is an indicator that is prepared by dissolving the solid in ethanol.
Since
Ka=[BCG−][H3O+][HBCG]When [BCG-] = [HBCG], then Ka = [H3O+].
If pH = 3.8
Ka= [H3O+] = -antilog pH = -antilog (3.8)
Ka= 1.58 ×10^-4
Answer:
Explanation:
In one of the process, energy is built up from scratch, in the other one, energy is liberated for use by an organism or body.
The first process deals with a metabolic reaction in which energy is liberated:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy
In the above process, energy is liberated when glucose combines with oxygen. The waste products are carbon dioxide and water. This process liberates heat energy which can be used to do work.
In the reverse process:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
This process stores energy in carbon chains as chemical energy. It is this energy that is released in the first process.
Therefore, we can see that the first process liberates energy and the reverse process stores energy.
