the answer was S=d/t = 60/10 = 6m/s, I'm sorry for getting it wrong the first time I answered ^^
Molar mass of Al₂(CrO₄)₃ is the mass of one mole of Al₂(CrO₄)₃
one mol =weight of Al₂(CrO₄)₃ in g /one mol
1) We have to find the atomic mass (in a periodic talbe) of the following elements (in u)
Al=26.98 u
Cr=52 u
O=16 u
2)We calculate the atomic weight of Al₂(CrO₄)₃
Atomic weight (Al₂(CrO₄)₃)=2(26.98 u)+3[(52 u)+4(16 u)]=
=53.96 u+3(52 u+64 u)=53.96 u+3(116 u=53.96 u+348 u=401.96 u
3) One mole of (Al₂(CrO₄)₃ ) have 401.96 g of Al₂(CrO₄)₃
molar mass=401.96 g/ mol
answer: 401.96 g/ mol
Oooooooooooooooooooooooooooo
Explanation:
It is known that for first order reaction, the equation is as follows.
t = ![\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}](https://tex.z-dn.net/?f=%5Cfrac%7B2.303%7D%7BK%7D%20log%20%5Cfrac%7B%5BC_%7B4%7DH_%7B8%7D%5D_%7Bo%7D%7D%7B%5BC_%7B4%7DH_%7B8%7D%5D_%7Bt%7D%7D)
t = ?, K = rate constant = 79 1/s
Initial conc. of
= 1.68
Decompose amount of
= 52% of 1.68
= 
= 0.8736
= 0.87
Now,
= (1.68 - 0.87)
= 0.81
Therefore, calculate the value of t as follows.
t = ![\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}](https://tex.z-dn.net/?f=%5Cfrac%7B2.303%7D%7BK%7D%20log%20%5Cfrac%7B%5BC_%7B4%7DH_%7B8%7D%5D_%7Bo%7D%7D%7B%5BC_%7B4%7DH_%7B8%7D%5D_%7Bt%7D%7D)
= 
= 
=
s
= 
Thus, we can conclude that 0.00921 s will be taken for 52% of the cyclobutane to decompose.
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