Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.
Explanation:
To calculate the moles :

According to stoichiometry :
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 2 moles of
give = 2 moles of 
Thus moles of
give =
of 
Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.
Hey there!
C₆H₁₄ + O₂ → CO₂ + H₂O
Balance H.
14 on the right, 2 on the left. Add a coefficient of 7 in front of H₂O.
C₆H₁₄ + O₂ → CO₂ + 7H₂O
Balance C.
6 on the left, 1 on the right. Add a coefficient of 6 in front of CO₂.
C₆H₁₄ + O₂ → 6CO₂ + 7H₂O
Balance O.
2 on the left, 12 on the right. Add a coefficient of 6 in front of O₂.
C₆H₁₄ + 6O₂ → 6CO₂ + 7H₂O
Our final balanced equation: C₆H₁₄ + 6O₂ → 6CO₂ + 7H₂O
Hope this helps!
6.8 is the pH of the solution after 10 ml of 5M NaOH is added.
Explanation:
Data given:
Molarity of C6H5CCOH = 0.100 M
molarity of ca(c6h5coo)2 = 0.2 M
Ka = 6.3 x 10^-5
first pH is calculated of the buffer solution
pH = pKa+ log 10 ![\frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
pKa = -log10[Ka]
pka = -log[6.3 x10^-5]
pKa = 4.200
putting the values to know pH of the buffer
pH = 4.200 + log 10 
pH = 4.200 + 0.3
pH = 4.5 (when NaOH was not added, this is pH of buffer solution)
now the molarity of the solution is calculated after NaOH i.e Mbuffer is added
MbufferVbuffer = Mbase Vbase
putting the values in above equation:
Mbuffer = 
= 
= 0.01 M
molarity or [ A-] = 5M
pH = pKa+ log 10 ![\frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
pH = 4.200 + log 10 
pH = 4.200+ 2.69
pH = 6.8
Explanation:
frequency= velocity/ wavelength
= 340/1.36
=250
Answer If a ball goes into a pocket and bounces back on to the playing surface, it is not considered pocketed. If it is the 8-ball, it is not a win. If it is the cue ball, it is not a scratch.
Explanation: