Answer:
P₂ = 5000 KPa
Explanation:
Given data:
Initial volume = 2.00 L
Initial pressure = 50.0 KPa
Final volume = 20.0 mL (20/1000=0.02 L)
Final pressure = ?
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
50.0 KPa × 2.00L = P₂ × 0.02 L
P₂ = 100 KPa. L/0.02 L
P₂ = 5000 KPa
Answer:
67.4 % of C₉H₈O₄
Explanation:
To make titrations problems we know, that in the endpoint:
mmoles of acid = mmoles of base
mmoles = M . volume so:
mmoles of acid = 20.52 mL . 0.1121 M
mmoles of acid = mg of acid / PM (mg /mmoles)
Let's determine the PM of aspirin:
12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol
mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol
mass (mg) = 414.4 mg
We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g
We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
A neutral particle made of an electron and hole
Explanation:
Exciton
It is the combination of an electron and a hole ( hole refers to the vacancy of an electron ) . And , as both the electron and the hole have the same charge but the polarity is opposite , the combination will lead to a neutral compound , i.e. , Exciton have no charge and so neutral .
It is free to move in the nonmetallic crystal and since it charge less , it is difficult to detect it directly .
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
--------------------
Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
--------------------
Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**
