Answer:
Explanation:
<u>1. Equilibrium equation</u>
<u>2. Equilibrium constant</u>
The liquid substances do not appear in the expression of the equilibrium constant.
![k_c=\dfrac{[HBr(g)]^2}{[H_2]}=4.8\times 10^8M](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BHBr%28g%29%5D%5E2%7D%7B%5BH_2%5D%7D%3D4.8%5Ctimes%2010%5E8M)
<u>3. ICE table.</u>
Write the initial, change, equilibrium table:
Molar concentrations:
H₂(g) + Br₂(l) ⇄ 2HBr(g)
I 0.400 0
C - x +2x
E 0.400 - x 2x
<u>4. Substitute into the expression of the equilibrium constant</u>
<u>5. Solve the quadratic equation</u>
- 192,000,000 - 480,000,000x = 4x²
- x² + 120,000,000x - 48,000,000 = 0
Use the quadratic formula:
The only valid solution is x = 0.39999999851M
Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M
Answer:
HCl conc.= 6.0mol/L
Explanation:
From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,
Answer:
The scientific method is a system of reasoning. You can make reasonable explanations for something that you are observing in the world.
Explanation:
<h2><u>
Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
<h2><u>
Explanation:</u></h2>
<u></u>
PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
