Answer:
It will cost $68,620.5 to the club to add the grass turf to their field.
Explanation:
length of the field = l = 0.102 km = 0.102 × 1000 m
( 1km = 1000 m)
Width of the field w = 0.069 km = 0.069 × 1000 m
Area of the field , A= l × w

Cost of grass turfing = 
Cost of grass turfing on field of
:

It will cost $68,620.5 to the club to add the grass turf to their field.
Answer:
Where SiO2 has strong covalent bonds between molecules, SO3 only has weak van der waals intermolecular forces holding its molecules together. Since covalent bonds require more energy to overcome than van der waals, SiO2 requires a higher temperature than SO3 to melt, thus SiO2 has a higher melting point.
The answer is C which is PbSo
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.