Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
Answer:
The answer to your question is an acid base reaction
Explanation:
A single replacement reaction is a reaction in which one metal replaces the cation of a compound. The reaction of this problem is not of this type because here the reactants are compounds no single elements.
A decomposition reaction is a reaction in which one compound decomposes into two or more products. This is not the answer to this question because in this reaction there are two reactants not only one.
A synthesis reaction is a reaction in which two reactants form only one product. The reaction of this problem is not of this type because there are two products not only one.
An acid-base reaction is a kind of double replacement reaction. In some acid-base reactions, there is an interchange of cations and anions like is shown in this reaction.
2.52 answer=c. That is the pH.
Answer:
Explanation:
Given that:
From above:
To predict the effect of the addition of Br₂(g);
The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr
The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.
Answer:
1. 
was the 
value calculated by the student.
2. 
was the 
of ethylamine value calculated by the student.
Explanation:
1.
The 
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_a=\frac{[As^-][H^+]}{[HAs]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BAs%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHAs%5D%7D)
:
was the value calculated by the student.
2.
The value of ethylamine = 11.87
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BC_2H_5NH_3%5E%7B%2B%7D%5D%5BOH%5E-%5D%7D%7B%5BC_2H_5NH_2%5D%7D)
:
was the of ethylamine value calculated by the student.
