Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>
Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
Best regards!
Answer:
Explanation:
The relation between equilibrium constant and Ecell is given below .
E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K
E⁰cell = 1.46 - 1.21 = .25 V
n = 2
Putting the values
.25 = (8.314 x 298 lnK) / (2 x 96485 )
lnK = 19.47
K = 2.85 x 10⁸
2 )
Change in free energy Δ G
Δ G ⁰ = nE⁰ F
n = 4
E⁰ = .4 + .83 = 1.23 V
Δ G ⁰= 4 x 1.23 x 96485
= 474706 J / mol
3 )
E⁰cell = (RT / nF ) lnK
n = 2
1.78 = 8.314 x 298 lnK / 2 x 96485
lnK = 138.638
K = 1.62 x 10⁶⁰
B because it’s got the least connections
