Answer : The final temperature would be, 791.1 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 265 kJ/mol = 265000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B4%5Ctimes%20K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B265000J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B733K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the final temperature would be, 791.1 K
Mg + 2️⃣H20 ➡️ Mg(OH)2 + H2
The molarity of a solution that contains 20.0grams of sodium hydroxide in 200. ml of solution is 2.3 M.
<h3>What is Molarity? </h3>
Molarity is defined as the ratio of number of moles of solute to the volume of the solution.
Molarity = number of moles of solute/ Volume of solution
Given,
given mass of sodium hydroxide= 20 g
Volume of the solution = 200 mL.
Firstly, we will calculate the moles of sodium hydroxide.
<h3>What is Mole? </h3>
Mole is defined as the ratio of given mass of substance to the molar mass of substance.
Mole = given mass/ molar mass
Molar mass of sodium hydroxide = 40 g
Mole = 20/40
= 0.5 mol
Molarity = (0.5/ 200) × 1000 = 2.5M
Thus, we calculated that the molarity of a solution that contains 20.0grams of sodium hydroxide in 200. ml of solution is 2.3 M.
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2:7 ratio of ethane to O2 = 15:x 9solve for x)
x=52.5 mol O2
