4Al(s) + 30₂(g) → 2Al2O3 (s)

One of the products of a fermentation reaction is

Alex73 [517]

There are two types of fermentation aerobic and anaerobic. Aerobic are those that need oxygen to ferment while anaerobic are those who do not need oxygen. Products of fermentation are usually water, ethanol, lactic acid and carbon dioxide. Other products are acetone and butyric acid.

5 0

3 years ago

The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compre

marysya [2.9K]

Answer:

0.0126 moles are released

Explanation:

Using Henry's law, where the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid:

S = k×P

Where S is solubility (5.6x10⁻⁴molL⁻¹), k is Henry's constant and P is partial pressure (0.80atm)

Replacing:

5.6x10⁻⁴molL⁻¹ / 0.80atm = 7x10⁻⁴molL⁻¹atm⁻¹

Thus, with Henry's constant, solubility of N₂ when partial pressure is 3.8atm is:

S = 7x10⁻⁴molL⁻¹atm⁻¹ × 3.8atm

S = 2.66x10⁻³molL⁻¹

Thus, when the deepd-sea diver has a pressure of 3.8amt, moles dissolved are:

6.0L × 2.66x10⁻³molL⁻¹ = 0.01596 moles of N₂

At the surface, pressure is 0.80atm and solubility is 5.6x10⁻⁴molL⁻¹, moles dissolved are:

6.0L × 5.6x10⁻⁴molL⁻¹ = 3.36x10⁻³mol

Thus, released moles are:

0.01596mol - 3.36x10⁻³mol = 0.0126 moles are released

8 0

3 years ago

how many valence electrons are found in most atoms of noble gases use this information to explain why the noble gases are unreac

kvasek [131]

8 valence electrons are found in noble gases. Since noble gases have 8 valence electrons that means they have a full shell - which makes them unreactive

3 0

3 years ago

Match each property to the appropriate subatomic particle.

sladkih [1.3K]

Proton = +1
neutron = 0
electron = -1

4 0

3 years ago

I just need help on c-g. The whole question is there in case anyone needs it.

Elza [17]

Answer:

(c) 0.11; (d) -24.5 kJ·mol⁻¹; (e) See below; (f) Kc increases;

(g) No effect on ΔH

Step-by-step explanation:

(c) Kc at 125 °C

In Part (a) one molecule of XY dissociated into one X and one Y.

XY ⇌ X + Y

I: 10 0 0

C: -1 +1 +1

E: 9 1 1

Kc = {[X][Y]}/[XY] = (1 × 1)/9 = ⅑ = 0.11

(d) ΔH

ΔH is a constant that is characteristic of the reaction.

ΔH = -24.5 kJ·mol⁻¹

(e) Effect of temperature on concentrations

If ΔH is negative, the reaction is exothermic.

Heat is a product of the reaction, so we can write the equation as

XY ⇌ X + Y + heat

If we lower the temperature, we are removing heat from the system.

Le Châtelier's Principle states that if you apply a stress to a system at equilibrium, it will respond by trying to relieve the stress.

We applied a stress by removing heat, so the system responds by producing more heat. The position of equilibrium moves to the right, and more products will form.

The diagram might look like the one below.

(f) Effect of temperature on Kc

Kc = [Products]/[Reactants]

We are increasing [Products] and decreasing [Reactants].

If you increase the numerator and decrease the denominator, you increase the value of the quotient.

The value of the equilibrium constant increases when the temperature decreases.

(g) Effect of temperature on ΔH

Decreasing the temperature has no effect on ΔH, because the enthalpies of the reactants and products are properties of the substances themselves. They do not depend on the temperature.

7 0

2 years ago