Answer: 
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 = 262 kJ/mol = 262000J/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{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)


Therefore, the value of the rate constant at 775.0 K is 
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Continuous spectrum and line spectrum are two types of spectra; their main difference is that continuous spectrum contains no gaps whereas line spectrum contains many gaps.
Answer:
619°C
Explanation:
Given data:
Initial volume of gas = 736 mL
Initial temperature = 15.0°C
Final volume of gas = 2.28 L
Final temperature = ?
Solution:
Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)
Initial temperature = 15.0°C (15+273 = 288 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 2.28 L × 288 K / 0.736 L
T₂ = 656.6 L.K / 0.736 L
T₂ = 892.2 K
K to °C:
892.2 - 273.15 = 619°C
B) Morrison and Franscioni's research done to create the Frisbee.