The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
Answer:
The importance of significant figures
As stated before, it is important within the science fields that you are not more precise or accurate than the least accurate or precise number. In science, it is generally agreed upon that the last number digit in any figure is filled with uncertainty.
Explanation:
ANS: density = 13.41 g/ml
Density (d) of a substance is the mass (m) occupied by it in a given volume (v).
Density = mass/volume
i.e. d = m/v
m = (d) v -----(1)
The given equation from the plot of weight vs volume is :
y = 13.41 x ----(2)
Based on equations (1) and (2) we can deduce that the density of the metal is 13.41 g/ml
Answer:
7200 kPa
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas
Make P' the subject of the equation
P' = PVT'/V'T.............. Equation 2
Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K
Substitute these values into equation 2
P' = (900×8×600)/(2×300)
P' = 7200 kPa
Answer:
Food, safe water, housing, education, and health care are examples of basic needs. Wants are desires that can be satisfied by goods and services.Secondary needs are the desires and wants that become important when primary needs are satisfied. Definition (2): Secondary needs are connected with the desire for satisfaction and pleasure of the human being: designer articles or furniture, the high-tech and the newest cell phone, jewelry, a luxury car, etc.
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