52° 31' 12.0288"N and 13° 24' 17.8344" E. What continent is it on?

Which is an example of a beneficial mutation?

nata0808 [166]

Answer:

D

Explanation:

7 0

3 years ago

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Determine the value of the equilibrium constant, Kgoal, for the reaction

hodyreva [135]

Answer:

Part A

$K_{goal}= 3.22\times 10 ^{-66}$

Part B

$K_{goal}= 2.26\times10^{-21}$

Explanation:

Part A:

The given equation is not balanced. So, initially let's balance the equation by taking 24 moles of each of the reagent and NO.

24N₂ + 24H₂O ⇌ 24NO + 12N₂H₄

Now, simplify the equation by dividing both sides by 12. The final balanced equation is the following

2N₂ + 2H₂O ⇌ 2NO + N₂H₄

The above-balanced equation can be solved algebraically to obtain the required Kgoal value.

Adding given equations 1, 2 and 3 we obtain the required equation.

When the equations are added, the equilibrium constants of each equation are multiplied. Mathematically it can be represented as,

$K_1 \times K_2 \times K_3 = K_{goal}$

$K_{goal}= 4.10 \times 10^{-31} \times 7.40 \times 10^{-26} \times 1.06 \times 10^{-10}$

$K_{goal}= 3.22\times 10 ^{-66}$

Part B:

The required equation is balanced, Now

Let.

P₄(s)+6Cl₂(g) ⇌ 4PCl₃(g), K₁=2.00×10¹⁹ ------------------------------------ (a)

PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), K₂=1.13×10⁻² --------------------------------------- (b)

By multiplying equation 2 by 4 and subtracting equation 1 from it, we get

4PCl₅(g) ⇌ P₄(s)+10Cl₂(g)

The Kgoal for the above equation is the product of four times K₂ and inverse K₁ according to the applied operation. Mathematically,

$K_{goal}= 4K_2 \times \frac{1}{K_1}$

$K_{goal}= 4(1.13\times10^{-2}) \times \frac{1}{2.00\times10^{19}}$

$K_{goal}= 2.26\times10^{-21}$

5 0

3 years ago

What is the process of scientific inquiry and why it is not a rigid sequence of steps?

k0ka [10]

Math and hint alots

4 0

4 years ago

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Identify each process numbered 1-6

Vedmedyk [2.9K]

Answer:

1 = Melting

2 = Freezing

3 = Sublimation

4 = Deposition

5 = Condensation

6 = Boiling

Explanation:

7 0

3 years ago

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.799-(-0.762)=1.561V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago