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3 years ago

What does the group number on the periodic table tell you about the electron configuration of a

Chemistry

1 answer:

jarptica [38.1K]3 years ago

4 0

Answer:

The group number in the periodic table represents number of valence electrons of the elements in a certain group.

Explanation:

There are s, p, d, and f blocks, which you can see in periodic table

The s-block and p-block together are usually considered main-group elements, the d-block corresponds to the transition metals, and the f-block encompasses nearly all of the lanthanides (like lanthanum) and the actinides (like actinium)

There are three main principles, which may useful for you:

The Pauli exclusion rule basically says that at most, 2 electrons are allowed to be in the same orbital.
Hund’s rule explains that each orbital in the subshell must be occupied with one single electron first before two electrons can be in the same orbital.
The Aufbau process describes the process of adding electron configuration to each individualized element in the periodic table.

Hope this helps!

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As you have seen in this lab, the density of water is near 1 g/cm3. Anything with a density lower than this will float in water,

hammer [34]

The third answer is the one you want. You have to have an adjustable density. All other things being equal, if the tanks you use for holding just water when filled with water will let the sub sink, because the sub is made of a dense metal like iron or steel.

If on the other hand you fill these tanks with air, the net density will be below one and the sub will rise.

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2 years ago

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Aluminum reacts with chlorine gas to produce aluminum chloride according to the following equation. Al + Cl2 → AlCl3 Which of th

anzhelika [568]

<h3><u>Answer</u>;</h3>

B. 3/2

<h3><u> Explanation;</u></h3>

Balance the chemical equation

2Al + 3Cl2 → 2AlCl3

We want to convert moles of AlCl3 to moles of Cl2

The conversion factor is 2 mol AlCl3/3 mol Cl2.

We choose the one that makes the units cancel:

x mol AlCl3 x (3 mol Cl3)/(2mol AlCl3) = x mol Al

The fraction for the molar ratio is 3/2.

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3 years ago

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Which would be the best way to represent the concentration of a 1. 75 M K2CrO4 solution? 1. 75% K2CrO4 [K2CrO4] (K2CrO4) K2CrO4,

tekilochka [14]

The amount of the solute present in the given solution is called the concentration. The best way to represent the concentration of the solution is $\rm [K_{2}CrO_{4}].$

<h3>What is molar concentration?</h3>

Molar concentration is the molarity of the solution that is the measure of the concentration of the solute dissolved in the solution.

The formula for calculating molar concentration is given as,

$\rm Molarity = \dfrac{\text{moles of solute}}{\text{Solution in L}}$

The concentration of any substance is represented in the square bracket like $\rm [X]$ or $\rm [K_{2}CrO_{4}].$

Therefore, option B. $\rm [K_{2}CrO_{4}]$ is the representation of the concentration.

Learn more about the molarity here:

brainly.com/question/1532164

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2 years ago

If 500ml of H₂ gas at 600mmHg and 400ml of CO₂ gas at 700mmHg are mixed in a 1 lit vessel, find the total pressure of mixture of

seraphim [82]

Answer:

Total pressure increased

Explanation:

When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence number of collision increases therefore pressure also increases because number of collision increases.

Total pressure increases.

6 0

2 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

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3 years ago