Answer:
Explanation:
Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V
Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V
Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .
So reaction in the combined cell will be
2Fe⁺³ + Fe = 3Fe⁺²
cell potential = .77 - ( - .44 )
= 1.21 V .
Answer:
The correct option is b. an amino-terminal signal
Explanation:
A polypeptide that will eventually fold to become an ion channel protein, it means a kind of integral membrane protein, has an amino terminal signal that indicates its delivery to endoplasmic reticulum (ER) and then to the membrane. This type of signal usually consist in a nucleus of 6 to 12 aminoacids and one or more basic aminoacids. Once the polypeptide enters the ER, this signal is removed.
Answer:
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
Less than the concentration of Pb2+(aq) in the solution in part ( a )
Explanation:
From the question:
A)
We assume that s to be the solubility of PbI₂.
The equation of the reaction is given as :
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹
[Pb²⁺] = s
Then [I⁻] = 2s
![K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BI%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%20%3D%20s%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%204s%5E%7B3%7D%5C%5Cs%5E%7B3%7D%20%3D%20%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%5C%5C%5C%5Cs%20%3D%5Cmathbf%7B%20%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5Ctext%7BThe%20mathematical%20expressionthat%20can%20be%20used%20to%20determine%20the%20value%20of%20%20%7D%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
B)
The Concentration of Pb²⁺ in water is calculated as :
![\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7B7%2A10%5E%7B-9%7D%7D%7B4%7D%7D%7D)
![\mathbf{s} =\sqrt[3]{1.75*10^{-9}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%7D%20%3D%5Csqrt%5B3%5D%7B1.75%2A10%5E%7B-9%7D%7D)
The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI
The equilibrium constant:
![K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5BPb%5E%7B2%2B%7D%7D%5D%5BI%5E-%5D%5E2%20%5C%5C%20%5C%5C%20K_%7Bsp%7D%20%3D%20s%2A%281.0%2A2s%29%5E2%20%3D7%2A1.0%5E%7B-9%7D%20%5C%5C%20%5C%5C%20s%20%3D%207%2A10%5E%7B-9%7D%20%5C%20%5C%20%20m%2FL)
It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
In order to calculate the number of atoms, we must first know the number of moles present. And
moles = (mass present) / (molecular mass)
Therefore, the moles of Mg present are
170 / 24 = 7.08
The number of atoms in a mole of substance is given by Avagadro's Number which is 6.02 x 10^23
Since there are 7.08 moles, there are:
7.08 * 6.02*10^23
= 4.26 * 10^24 atoms
