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3 years ago

What is the difference between a 1s orbital and a 2s orbital is that

Chemistry

1 answer:

iVinArrow [24]3 years ago

4 0

Answer:

The main difference is their energy level, 2s orbital is higher than 1s orbital.

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Select the correct answer.

Artist 52 [7]

I think c cause if he is in a relationship that’s abusive he needs to speak up before it goes way worst.

3 0

3 years ago

The energy required to ionize a mole of potassium ions is 419 kJ/mol . What is the longest wavelength of light capable of this i

liraira [26]

First divide the ionization energy by Avogadro's number to get the energy per atom of potassium;
419 kj/mol / 6.023 x 10²³
= 4.19 x 10⁵ / 6.023 x 10²³ = 6.96 x 10⁻¹⁹
E = hc/λ
where lambda (λ) is the wavelength, h is Planck's constant, c is the speed of light
E = 6.96 x 10⁻¹⁹ j/atom
h =6.63x10⁻³⁴ Js
c = 3 x 10⁸ m/s
λ = ?
λ = hc/E = (6.63x10⁻³⁴ x 3 x 10⁸ ) / 6.96 x 10⁻¹⁹ = 285.8nm = 286nm
The longest wavelength of light capable of this ionization is 286nm.

3 0

3 years ago

How many moles of CO are present in 35.88 L of the gas?

Pepsi [2]

Answer:

1.602 moles CO

Explanation:

To convert from liters to moles, divide by 22.4:

35.88 L / 22.4 = 1.602 moles CO

7 0

3 years ago

Which ligand binds tightest? ligand A, with a dissociation constant ( Kd ) of 10−9 M ligand D, with a percent occupancy of 80% a

Delvig [45]

Answer:

The answer will be Ligand A with a dissociation constant (Kd) of $10^{-9}$ M

Explanation:

When the dissociation constant in the ligand is small (in order of nano) ( $10^{-9}$ ) it will be more tied. Due to a dissociation constant measures how much a ligand can be able to be separated from the protein so if the number is small it means that the ligand is highly binded to the protein.

On the other hand, the occupancy percentage of the ligand does not imply binding. Conversely, a High-affinity ligand binding with the proteins implies that a relatively low concentration of a ligand is adequate to occupy the maximum ligand-binding site.

6 0

3 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

Answer:

For (a): The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

For (b): The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

For (c): The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

For (d): The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

For (e): The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago