Answer:
Explanation:
Step 1: Data given
The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K
The equilibrium concentration of Cl2(g) is 0.233 M
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3: The initial concentration
[PCl5]= Y M
[PCl] = 0M
[Cl2] = 0M
Step 4: Calculate the concentration at equilibrium
[PCl5] = Y + X M = Y - 0.233 M
[PCl]= XM = 0.233 M
[Cl2]= XM = 0.233 M
Step 5: Define Kc
Kc = [Cl2]* [PCl3] / [PCl5]
4.76 * 10^-4 = 0.233² / (Y -0.233)
0.000476 = 0.05429 / (Y - 0.233)
Y - 0.233 = 0.05429 / 0.000476
Y - 0.233 = 114.05 M
Y = 114.283 M = the initial concentration
The concentration of PCl5 at the equilibrium is 114.05 M
Answer:
Explanation:
Which of the following elements would lose an electron easily?
In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron
Answer:
Air Pressure and Wind direction
Explanation:
Just took the test on e2020
Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=
V
nRT
. Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}
mol⋅K
atm⋅L
:
P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=
V
nRT
=
15.0
L
(0.33
mol
)(0.0821
mol⋅K
atm⋅
L
)(248.15
K
)
=0.4482atm
In conclusion, the pressure of this gas is P=0.4482 atm.
Reference:
Chang, R. (2010). Chemistry. McGraw-Hill, New York.
