Answer:
Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]
You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76
(a). pH = 4.76 + log [0.13]/[0.10]
= 4.76 + 0.11
= 4.87
(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt
pH = 4.76 + log [0.15]/[0.08]
= 4.76 + 0.27
= 5.03
Answer:
P = 83.16 Watt
Explanation:
Given data:
Mass of box = 2 Kg
velocity of box = 29.7 m/s
Time = 21 sec
Power = ?
Solution:
Formula:
Power = work done/time
Now we will calculate the work done.
W = force × distance ...... (1)
d = speed /time
d = 29.7 m/s× 21 s
d = 623.7 m
Force:
acceleration = 29.7 m/s / 21 s
a = 1.4 m/s²
Force = m×a
Force = 2 kg×1.4 m/s²
Force = 2.8 N
Now we will put the values in equation 1.
W = F × d
W = 2.8 N ×623.7 m
W = 1746.36 Nm
Now for power:
P = W / time
P = 1746 Nm / 21 s
P = 83.16 Watt
your answer is 0.00833M the volume was converted into liters
NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)
Using the molarity equation, we can find the number of moles of HCl that reacted:
molarity=mol soluteL soln
mol solute=(molarity)(L soln)
mol HCl=(0.105molL)(0.0250L)=0.00263 mol HCl
(volume converted to liters)
Now, using the coefficients of the chemical reaction, we can determine the number of moles of NaOH that reacted:
0.00263mol HCl(1lmol NaOH1mol HCl)=0.00263 mol NaOH
Lastly, we'll use the molarity equation (using given volume of NaOH soln) again to determine the molarity of the sodium hydroxide solution:
molarity=mol soluteL soln
Answer:
There is a basic difference, because mass is the actual amount of material contained in a body and is measured in kg, gm, etc. Whereas weight is the force exerted by the gravity on that object mg. Note that mass is independent of everything but weight is different on the earth, moon, etc.
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
