Answer:
breathing
Explanation:
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P1V1=P2V2
(600/760)(.200)=(780/760)x
x=.1538L or 153.8mL
Answer:
34.21%
Explanation:
Given parameters:
Mass of ethylene glycol in the container = 130g
Mass of the liquid in total = 380g
Unkown:
Percentage of ethylene glycol in the liquid =?
Solution
To find the percentage of the ethylene glycol in liquid, we use the expression below:
Percentage of ethylene glycol in liquid =
x 100
Percentage of ethylene glycol =
x 100
Percentage of ethylene glycol = 34.21%
Atomic number = 20 -------> Ca (calcium)
Calcium has four shells (1, 2, 3, 4) and two sublevels ( s, p ).
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Answer : The activation energy for the reaction is, 51.9 kJ
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 295 K
= rate constant at 305 K = 
Ea = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 295 K
= final temperature = 305 K
Now put all the given values in this formula, we get:
![\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B2K_1%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B295K%7D-%5Cfrac%7B1%7D%7B305K%7D%5D)

Therefore, the activation energy for the reaction is, 51.9 kJ