What is the answer to the multiplecation problem -5 (-14) (-2)=

What is the correct equilibrium constant expression for equation P2(g)

sertanlavr [38]

Answer:

k = [F2]² [PO]² / [P2] [F2O]²

Explanation:

In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:

P2(g) + 2F2O(g) ⇄ 2PO(g) + 2F2(g)

The equilibrium constant, k, is:

k = [F2]² [PO]² / [P2] [F2O]²

6 0

3 years ago

Abuse of time and temperature can occur in each of the following ways except: The items are not cooked to the proper internal te

exis [7]

Holding items in the freezer for too long is not an example of abuse of time and temperature.

<h3>What is abuse of time and temperature?</h3>

Abuse of time and temperature has to do with a situation in which time is used in an improper way or food is not held at the correct temperature ultimately leading to food poisoning.

Holding items in the freezer for too long is not an example of abuse of time and temperature.

Learn more about temperature: brainly.com/question/7510619

5 0

3 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

<u>Answer:</u> The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

<u>Explanation:</u>

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

<u>Reduction half reaction:</u> $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

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:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

7 0

3 years ago

What type of energy is stored in the earths crust and produces earthquakes when It is released

mylen [45]

Heat energy because the energy in seismic waves that causes the ground to shake is heat energy. Heat energy associated with friction on the fault slip surface

Please mark me as braunliest

3 0

3 years ago