Answer:
seafloor spreading
Explanation:
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is carried out in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm3. If the equilibrium conversion is found to be 60%, (a) What is the equilibrium constant, KC if the reaction is a gas phase reaction? (Ans.: Gas: KC = 0.328 dm3/mol)
Answer:
0.328 mol/dm³
Explanation:
We have
I started this calculation from Rate's law.
Remember equilibrium constant has been given to be 60%
Our interest is Kc, that is the equilibrium constant.
Ca = 4(1-0.6)/1+(-0.5*0.6)
= 4-2.4/1-0.3
= 1.6/0.7
= 2.2857
Cb = 4x0.6/2(1+(-0.5*0.6))
= 2.4/2(0.7)
= 2.4/1.4
= 1.7143
Kc = Cb/Ca²
= 1.7143/2.2857²
= 1.7143/5.2244
= 0.328 mol/dm³
I have added an attachment showing earlier stages to the final answer
What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)
Answer:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Explanation:
Hello.
In this case, for the undergoing chemical reaction:
[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]
We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
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Calculate the volume of the gas, in liters, if 1.75 mol has a pressure of 1.28 atm at a temperature of -7 ∘C
Answer:
A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.
You need to apply the ideal gas law PV=nRT
You have the pressure, P=1.01 atm
you have the volume, V = 2.21 L
The ideal gas constant R= 0.08205 L. atm/ mole.K at 273 K
find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)
n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and
L=1.23 L
P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L
= 1.994 atm
Explanation:
the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?
Answer:
768g
Explanation:
We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.
The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram were in the original sample.
What is half life?The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.
The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.
ln P = kt + C
P = amount of I-137 at time t
C = constant
k = 1/time
t = time
1st condition:
P = Po, t = 0 days
2nd condition: (half-life)
P = 0.5Po, t = 8.07 days
3rd condition:
P = 25 grams, t = 40.35 days
Po = 800 grams
mass of I-137 = 800 gram
Therefore, 800 gram were in the original sample.
To learn more about half life, here:
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The following are placed in a beaker weighing 39.457 g:
2.689 g of NaCl, 1.26 g of sand and 5.0 g water
What is the final mass of the beaker?
Answer:
48.4 g
Explanation: