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Alex787 [66]
3 years ago
12

Will mark as Brainliest.

Chemistry
1 answer:
sertanlavr [38]3 years ago
8 0
<span>Atomic radius generally increases as we move 
</span><span>down a group and from right to left across a period.</span>
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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
Diethyl ether is produced from ethanol according to the following equation:
juin [17]
The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of  CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3 

% yield = .775 = actual yield / 264.66


actual yield = 205.11 g CH3CH2OCH2CH3
6 0
3 years ago
The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres
Dima020 [189]

Answer:

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures  of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as K_{p}

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

Partial pressures at equilibrium:

p^o_{H_2O}=0.070 atm

p^o_{H_2}=0.0035 atm

p^o_{O_2}=0.0025 atm

The equilibrium constant in terms of pressures is given as:

K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}

K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

5 0
3 years ago
Which type of radioactive emission has a positive charge and weak penetrating power?
Wewaii [24]
1. alpha particle
alpha particles are positively charged helium nuclei with weak penetrating power
5 0
3 years ago
Over the last several decades, scientists have addressed the problem of nonrenewable natural resources such as fossil fuels. Hum
Varvara68 [4.7K]

The answer is D

Hope this helped

6 0
3 years ago
Read 2 more answers
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