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
Answer: The transition elements are in the d-block, and in the d-orbital have valence electrons. They can form several states of oxidation and contain different ions. Inner transition elements are in the f-block, and in the f-orbital have valence electrons.
Explanation:
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
Density can be calculated using the following rule:
density=mass/volume
therefore,
volume=mass/density
we have mass=0.451g and density=0.824g/ml
substituting in the above equation, we can calculate the volume as follows:
volume = 0.451/0.824 = 0.547 ml
Answer:
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
Explanation:
Step 1: Data given
Molar mass helium = 4.0 g/mol
Molar mass O2 = 32 g/mol
Step 2: Graham's law
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5
Rate of escape for He = 1/(4.0)^0.5 = 0.5
Rate of escape for O2 = 1/(32)^0.5 = 0.177
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
