Is this a scientific model? Use complete sentences to explain why or why not

The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

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Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

3 0

2 years ago

If the oxidation number of an atom decreases, the atom has undergone oxidation.

Bumek [7]

Oxidation is a process when an atom, a group of atoms, or an ion loses electrons.

For example,

Cu⁰ -------> Cu⁺² + 2e⁻. This is an oxidation process.

As we can see an oxidation number was 0, and it became +2. The oxidation number is increased. Oxidation number of the atom, when an atom has undergone oxidation, increases.

So, answer is b.False.

8 0

2 years ago

EARTH SCIENCE: PLEASE HELP ME !!! Which tools do astronomers use to determine the effects of collisions between galaxies? Select

topjm [15]

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Astronomy deals with the study of outer space. The study focuses on planets and other celestial bodies. Generally telescopes are used to study outer space. The particular type of telescope that is used depends on the need.

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Learn more: brainly.com/question/556195

6 0

1 year ago

Which acid and base react to form water and sodium sulfate?

joja [24]

Answer:

acid: sulphuric acid (H2SO4)

base: sodium hydroxide (2NaOH)

Explanation:

3 0

2 years ago

Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c

KiRa [710]

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps

Initial x

React s

Eq x - s s 2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s = ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

3 0

2 years ago