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).
<span>ideal gas law is: PV = nRT
P = pressure (torr) = 889 torr
V = volume (Liters) = 11.8 L
n = moles of gas = 0.444 mol
R = gas constant = 62.4 (L * torr / mol * k)
solve for T (in kelvin)
T = PV/nR
T = (889*11.8)/(.444*62.4)
T = 378.6 K
convert to C (subtract 273)
T = 105.6 deg C</span>
Answer:
just answer this and you will have yours
Explanation:Find the area of a circle with a diameter of \color{green}{16}16start color green, 16, end color green.
Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter your answer as a decimal.
Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T
Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J
T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K
Answer:
1 M
Explanation:
Magnesium chloride will furnish chloride ions as:
Given :
Moles of magnesium chloride = 0.20 mol
Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:
Moles of chloride ions by magnesium chloride = 0.40 moles
Potassium chloride will furnish chloride ions as:
Given :
Moles of potassium chloride = 0.10 moles
Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:
Moles of chloride ions by potassium chloride = 0.10 moles
Total moles = 0.40 + 0.10 moles = 0.50 moles
Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)
Concentration of chloride ions is:
<u>
The final concentration of chloride anion = 1 M</u>