Answer:
12 mi/h
Explanation:
Step 1: Given data
Step 2: Convert "d" to miles
We will use the conversion factor 1 mi = 1.60934 km.
6 km × 1 mi/1.60934 km = 3.7 mi
Step 3: Convert "t" to hours
We will use the conversion factor 1 h = 60 min.
19 min × 1 h/60 min = 0.32 h
Step 4: Calculate the average speed of the runner (s)
The speed is equal to the quotient between the total distance and the time elapsed.
s = d/t
s = 3.7 mi/0.32 h = 12 mi/h
Positive unless your mean to them then everyones negative towards you.
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Answer is 16
carbon valence = 4
oxygen valence = 6 but carbon dioxide have 2 oxygen so = 12
12 + 4 = 16
so carbon dioxide have 16 electron
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\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 of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol
![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
