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.
Desert plants commonly have small wax coated leaves to prevent moisture and water from evaporating. It is an important feature in desert plants to adapt to the arid climate. The wax on the leaves also protect the leaves from the chilling temperature of the desert at night.
Answer:
The answer is 1.61 × 10²³ atoms
Explanation:
To determine number of atoms, we will use the formula below
Number of atoms = number of moles (n) × avogadro's constant (6.02 x 10²³)
n was not provided, hence we will solve for n
n = mass/ molar mass
molar mass of carbon monoxide, CO (where C is 12 and O is 16) is 12 + 16 = 28
mass was provided in the question as 7.48
n = 7.48/28
n = 0.267
Hence,
number of atoms = 0.267 × 6.02 x 10²³
= 1.61 × 10²³ atoms
