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2 months ago

Which of the following ions is formed when a base is dissolved in a solution? H+ O− OH− SO42+

Chemistry

1 answer:

eduard [2.7K]2 months ago

3 0

Result:

OH⁻

Explanation:

A base is characterized as a substance that reacts with water to produce an excess of hydroxide ions, OH⁻, within an aqueous environment.

For instance, Sodium hydroxide (NaOH)

and Potassium hydroxide (KOH)

Typically, bases consist of specific metallic oxides, metallic hydroxides, and aqueous ammonia.

On the other hand, an acid is known as a proton donor, H⁺, and it reacts with water to release surplus hydroxonium ions in an aqueous solution.

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Which factors are needed for organisms to live earth

Alekssandra [3086]

Answer: sunlight, water, air, habitat, and food.

Explanation: All living organisms rely on five fundamental necessities for survival: sunlight, water, air, habitat, and food.

3 0

2 months ago

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V 0 V0 for an enzyme‑

alisha [2963]

Answer:

The equation formulated by Michaelis-Menten is expressed as

v₀ = Kcat × [E₀] × [S] / (Km + [S])

in which,

Kcat denotes the experimental reaction rate constant; [S] signifies the concentration of the substrate, and

Km represents the Michaelis-Menten constant.

Explanation:

Refer to the attached image for an in-depth clarification

3 0

2 months ago

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

castortr0y [3046]

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

$Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)$

The half-reactions are:

Anode oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Cathode reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = $25^oC=273+25=298K$

n = electrons exchanged in oxidation-reduction = 2

$E^o_{cell}$ = standard electrode potential for the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction =?

$[Zn^{2+}]$ = concentration of Zn²⁺ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now substituting all known values into the equation, we arrive at:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

So, the resulting cell potential for this reaction is 0.50 V

5 0

1 month ago

Every single-celled organism is able to survive because it carries out *

castortr0y [3046]

Every unicellular organism prospers by executing metabolic activities.

Metabolic activities encompass the set of chemical reactions essential for sustaining life.

Explanation:

Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.

These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.

Organisms acquire the energy necessary for these activities through food consumption.