How do fungi create energy?

Consider the following system at equilibrium where H° = 16.1 kJ, and Kc = 6.50×10-3, at 298 K. 2NOBr(g) 2NO(g) + Br2(g)If the VO

Norma-Jean [14]

Answer:

1) The value of Kc : (C.) remains the same

2) The value of Qc : (A.) is greater than Kc

3) The reaction must : (B.) run in the reverse direction to reestablish equilibrium.

4) The number of moles of Br2 will : (B.) decrease

Explanation:

Value of concentration equilibrium constant Kc depends only on temperature. Since temperature remains constant, therefore, Kc remains constant. Decrease in volume means increase in pressure. Increase in pressure favors the side with less gaseous species. Hence, increase in pressure will favor the reverse reaction towards reactants.

7 0

3 years ago

Need help, I'm kinda panicking!

allochka39001 [22]

Answer:

hey I saw this on brainly and thought it will help u..

and this is not my answer I saw u needed help so I screenshot it..

have an amazing day..

and hope this is what ur looking for.

4 0

2 years ago

En una práctica experimental, para la obtención de cloruro cobaltoso, se hacen reaccionar 120 g de sulfuro cobaltoso de 60% de p

Ivenika [448]

Answer: D

Explanation:

Utilicé traductor de español para responder esta pregunta

4 0

3 years ago

Question: How will soap or alcohol affect the hydrogen bonds between different water molecules? Hypothesis: Materials: Penny, wa

Stella [2.4K]

Answer:

A soap or alcohol will interfere with the hydrogen bonding between water molecules making it weaker

Explanation:

A water molecule is polar as the oxygen atom is strongly electronegative and draws the electrons it shares with the hydrogen atoms in the water molecule to itself. This creates a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms. The attraction of neighboring oxygen to hydrogen atoms in a water molecule results in hydrogen bonding.

However, a detergent molecule which has a polar head and a hydrophobic tail, when added to water significantly interferes with the hydrogen bonding between water molecules. The polar head is attracted to water molecules whereas the hydrophobic tail of each molecule are not. The detergent molecules form clusters known as micelles with the polar heads attracted to water and the hydrophobic tails directed towards the center of the micelle, away from water molecules. This weakens the hydrogen bonding between water molecules

An alcohol contains a polar -OH group in its molecules. However, the hydrogen bonding between alcohol molecules are weaker than that in water because the other end of the alcohol molecule has a C-O bond which is less polar than the O-H bond. Therefore, when an alcohol is added to water, it weakens the hydrogen bonding in water.

This weakening of the hydrogen bonding in water also has the effect of lowering the surface tension of water.

7 0

3 years ago

Read 2 more answers

B. For the following questions, use the reaction NO2(g) N2(g) + O2(g), with ΔH = –33.1 kJ/mol and ΔS= 63.02 J/(mol·K).

Troyanec [42]

Answer:

I. Kindly, see the attached image.

II. The reaction is exothermic.

III. - 51.88 kJ/mol.

IV. The reaction is spontaneous.

Explanation:

I. Draw a possible potential energy diagram of the reaction. Label the enthalpy of the reaction.

Since the sign of ΔH is negative, the reaction is exothermic reaction.

In an exothermic reaction, the energy of the reactants is higher than that of the products.

Kindly see the attached image to show you the potential energy diagram of the reaction.

II. Is the reaction endothermic or exothermic? Explain your answer.

The reaction is exothermic reaction.

The sign of ΔH indicates wither the reaction is endothermic or exothermic one:

If the sign is positive, the reaction is endothermic.

If the sign is negative, the reaction is exothermic.

Herein, ΔH = - 33.1 kJ/mol, so the reaction is exothermic.

III. What is the Gibbs free energy of the reaction at 25°C?

∵ ΔG = ΔH - TΔS.

Where, ΔG is the Gibbs free energy change (J/mol).

ΔH is the enthalpy change (ΔH = - 33.1 kJ/mol).

T is the temperature (T = 25°C + 273 = 298 K).

ΔS is the entorpy change (ΔS = 63.02 J/mol.K = 0.06302 J/mol.K).

∴ ΔG = ΔH - TΔS = (- 33.1 kJ/mol) - (298 K)(0.06302 J/mol.K) = - 51.88 kJ/mol.

IV. Is the reaction spontaneous or nonspontaneous at 25°C?

The sign of ΔG indicates the spontaneity of the reaction:

If ΔG < 0, the reaction is spontaneous.

If ΔG = 0, the reaction is at equilibrium.

If ΔG > 0, the reaction is nonspontaneous.

Herein, ΔG = - 51.88 kJ/mol, so the reaction is spontaneous.

7 0

3 years ago