Answer:
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk
V/mL: 70.
c/mol·L⁻¹: 0.167
For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.
The equation is then
A + Ac₂O ⟶ B + junk
1. Moles of A

2. Moles of B
The molar ratio is 1 mol B:1 mol A
Moles of B = moles of A = 12 mmol = 0.012 mol

Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:

w is the mass flow
m is the mass of salt

v is the volume flow
C is the concentration





![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
Noble gases have complete valence electron shells
Answer:
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. ( B )
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not. ( C )
Explanation:
<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. and
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.
Given the relationship below,
Δ G = -nFE
E = emf of cell , G = free energy.
This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable