A scale and a ruler. The scale to measure the mass, and a ruler to measure the volume.
The final destination to where some of the electrons go to at the end of cellular respiration would be D. Oxygen. Assuming that this aerobic cellular respiration, the final electron acceptor is that of oxygen.
Answer:
2.5 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have two different values of V and P:
<em>P₁V₁ = P₂V₂
</em>
P₁ = 5.0 atm, V₁ = 3.5 L.
P₂ = 7.0 atm, V₂ = ??? L.
<em>∴ V₂ = P₁V₁/P₂ </em>= (5.0 atm)(3.5 L)/(7.0 atm) = <em>2.5 L.
</em>
Answer : The activation energy for the reaction is, 52.9 kJ/mol
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,
= initial temperature = 
= final temperature = 
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
Now put all the given values in this formula, we get:
![\log (\frac{25.0s^{-1}}{12.5s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{298.0K}-\frac{1}{308.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B25.0s%5E%7B-1%7D%7D%7B12.5s%5E%7B-1%7D%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B298.0K%7D-%5Cfrac%7B1%7D%7B308.0K%7D%5D)
Therefore, the activation energy for the reaction is, 52.9 kJ/mol
