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TiliK225 [7]
3 years ago
8

how do the outer shell electron configurations for ions of group 1,group 2, and group 15 how did the outer shell electron config

urations for ions of Group 1 group 2 and group 15 group 16 and group 17 elements compared with those of the noble gases
Chemistry
1 answer:
tangare [24]3 years ago
6 0

Explanation: Outer shell electronic configuration for Group 18 is :

\text{Group 18: }ns^2np^6 ( have 8 electrons in the outer shell)

The electronic configurations for other groups are:

\text{Group 1: }ns^1 ( have 1 electron in the outer shell}

\text{Group 2: }ns^2 ( have 2 electrons in the outer shell)

\text{Group 15: }ns^2np^3 ( have 5 electrons in the outer shell)

\text{Group 16: }ns^2np^4 ( have 6 electrons in the outer shell)

\text{Group 17: }ns^2np^5 ( have 7 electrons in the outer shell)

Comparing the electron configurations of group 18 to other groups.

Group 18 has fully filled outer shell, that is 8 electrons in the outer shell, while others are partially filled.


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Metamorphic rocks are formed by
marin [14]

Answer:

Heat and pressure.

Explanation:

Hope this helps!

7 0
3 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
2 years ago
Convert 90.23 kPa into atm
svlad2 [7]

Answer:

The answer is

<h2>0.89 atm </h2>

Explanation:

To convert from kPa to atm we use the conversion

101.325 kPa = 1 atm

If

101.325 kPa = 1 atm

Then

90.23 kPa will be

\frac{90.23}{101.325}  \\  =0.89050086...

We have the final answer as

<h3>0.89 atm</h3>

Hope this helps you

6 0
3 years ago
How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol
dsp73
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

<span>actual delta H = 0.34  moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>

<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
3 0
3 years ago
What is cation and anion for<br> CrS2
maksim [4K]
Cation - Cr 4+
Anion - S ^ 2-

Chromium sulfide
(:
4 0
1 year ago
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