Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
The chemical equation is
Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)
Answer:
12
Explanation:
In the right hand side of the equation, there are three compound which contains O2, which are;
Cu(NO3)2 , number of oxygen atoms =3*2 =6
2NO2, number of oxygen atoms = 2*2=4
2H2O, number of oxygen atoms =2*1=2
Total number of oxygen atoms on the right side of equation = 6+4+2 =12
<span>C + O2 → CO2
(8,376,726 tons) x (0.80) / (12.01078 g C/mol) x (1 mol CO2/ 1 mol C) x
(44.00964 g CO2/mol) = 24,555,054 tons CO2</span>
Physical......................................................
Answer:
13.53 kJ
Explanation:
The energy of a gas can be calculated by the equation:
E = (3/2)*n*R*T
Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.
E = (3/2)*3.5*8.314*310
E = 13,531.035 J
E = 13.53 kJ