<span>There
are a number of ways to express concentration of a solution. This includes
molarity and molality. Molarity is expressed as the number of moles of solute per volume of
the solution. MOlality is expressed as moles per kg solution.
5.25 mol H2SO4 / kg solution ( 1 kg / 1000 g ) ( 1.266 g / mL ) ( 1000 mL / 1L ) = 6.6 M H2SO4</span>
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
Explanation:
The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. Traditionally, the reaction uses potassium phthalimide. ... The alkylation of ammonia is often an unselective and inefficient route to amines. In the Gabriel method, phthalimide anion is employed as a surrogate of H2N−.
How much it has to drop and how heavy it is. Hope this is what you're looking for:)
This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
