Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
Hydrogen Peroxide slowly decomposing into water and oxygen increases the entropy of the universe. Any reaction of a system always increases the degree of the orderliness of the universe. Decomposition is one of the best examples of increasing the entropy. This is when substance is broken down further to elements or other compounds.
Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.
Answer:
252.68 K or -20.46 °C
Explanation:
According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".
Mathematically,
P₁ / T₁ = P₂ / T₂ ---- (1)
Data Given:
P₁ = 30.7 kPa
T₁ = 0.00 °C = 273.15 K
P₂ = 28.4 kPa
T₂ = <u>???</u>
Solving equation for T₂,
T₂ = P₂ T₁ / P₁
Putting values,
T₂ = 28.4 kPa × 273.15 K / 30.7 kPa
T₂ = 252.68 K or -20.46 °C
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
