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
M= #moles / L
4.35/.75 = 5.6
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
<h3>Answer:</h3>
Mass = 96.47 g
<h3>
Solution:</h3>
Data Given:
M.Mass = 28.97 g.mol⁻¹
Moles = 3.33 mol
Mass = ??
Formula Used:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 3.33 mol × 28.97 g.mol⁻¹
Mass = 96.4701 g
Rounding to four significant numbers,
Mass = 96.47 g
Answer is: sucrose is more soluble in water.
Solubility of sucrose (C₁₂H₂₂O₁₁) is about 2000 g in one liter of water (25°C) and solubility of lauric acid (C₁₂H₂₄O₂) is approximately 0,06 g approximately.
That is because sucrose has stronger intermolecular forces (hydrogen bond), Sucrose has more oxygen, more oxygen means more intermolecular bond with hydrogen.
