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dusya [7]
3 years ago
13

List all of the recessive traits you noticed in the picture

Chemistry
1 answer:
Otrada [13]3 years ago
3 0

Answer:

what picture

Explanation:

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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
6.
GuDViN [60]

Answer:

A phosphorus atom forms a P3- ion by gaining three electrons.

here

4 0
2 years ago
A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet
Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

m=1.67 g/mL\times 29.1 mL=48.597 g

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= \frac{70.5}{100}\times 48.597 = 34.261 g

Moles of perchloric acid = \frac{34.261 g}{100.46 g/mol}=0.3410 mol

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

Molarity=\frac{Moles}{Volume (L)}

=\frac{0.3410 mol}{0.250 L}=1.364 M

The concentration of the solution is 1.364 molar.

6 0
3 years ago
Give an example of one type of energy conversion (change to another form). Be sure to explain your example.
Daniel [21]
Potential to Kinetic Energy. 
When you a basketball in your hand that is potential energy. Then, when you bounce the ball that is kinetic energy. 
3 0
3 years ago
The emission spectrum of hydrogen shows discrete, bright, colored lines. Which characteristic of the Bohr model is best supporte
natali 33 [55]
The characteristic of the Bohr model that would best support his observation is this assumption: "The energy of the electron in an orbit is proportional to its distance from the nucleus. The further the electron is from the nucleus, the more energy it has." The discrete, bright, colored lines might represent the electrons and its distance from the nucleus. The lights are caused by the energy it has.
7 0
3 years ago
Read 2 more answers
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