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3 years ago

Consider the Bohr Model of Phosphorus below. Which statement is FALSE?

Chemistry

1 answer:

Rudik [331]3 years ago

7 0

Answer:

Explanation:

In the Bohr model of the atom, electrons are arranged in energy levels. The electrons in the lowest energy levels are nearest to the nucleus. An electron may move from a lower to a higher energy level by absorbing energy that is equal to the difference between the energies of the higher and lower energy level.

The potential energy of all electrons inside the atom have negative values and an electron which is infinitely far from the nucleus has an electrostatic potential energy of zero.

Energy is absorbed when an electron is removed from the atom (ionization). Hence the process is highly endothermic. Therefore it is false to say that "Energy is released when the electron is ejected from the atom."

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Magnesium chloride is a salt formed with ionic bonds between one magnesium ion and two chloride ions. Magnesium has two electron

katrin [286]

Answer:

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

Explanation:

Ionic compounds are compound formed from the transfer of electron(s). One atom of the element loses electron(s) while the other atom gains electron(s).

The compound Magnesium chloride is an ionic compound . The bond between an atom of magnesium and 2 atoms of chlorine is an ionic bonding.

The valency electron of magnesium is 2 electron , for the atom of magnesium to attain octet rule, it will easily lose it 2 electrons to the chlorine atoms.

The chlorine atom on the other hand has 7 valency electrons, to attain octet configuration it will most likely gain 1 electron to become stable.

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

7 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

3 years ago

When 8.0 grams of sodium hydroxide is dissolved in sufficient water to make 400. mL of solution, what is the concentration of th

nata0808 [166]

Answer:

The concentration of this sodiumhydroxide solutions is 0.50 M

Explanation:

Step 1: Data given

Mass of sodium hydroxide (NaOh) = 8.0 grams

Molar mass of sodium hydroxide = 40.0 g/mol

Volume water = 400 mL = 0.400 L

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 8.0 grams / 40.0 g/mol

Moles NaOh = 0.20 moles

Step 3: Calculate concentration of the solution

Concentration solution = moles NaOH / volume water

Concentration solution = 0.20 moles / 0.400 L

Concentration solution = 0.50 M

The concentration of this sodiumhydroxide solutions is 0.50 M

6 0

3 years ago

What do the symbols in the parentheses indicate

beks73 [17]

Answer:

Explanation:

Hope this helps

8 0

3 years ago

Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]

madreJ [45]

Answer:

$Mn^{2+}$ : - F . $[Ar]3d^{5}$

$Hg^{2+}$ : - G. $[Xe]4f^{14}5d^{10}$

$La^{3+}$ : - D. $[Xe]$

$Fe^{3+}$ : - F. $[Ar]3d^{5}$

$Ag^{+}$ : - C. $[Kr]4d^{10}$

$Co^{3+}$ : - E. $[Ar]3d^{6}$

Explanation:

The electronic configuration of the element Mn is:-

$[Ar]3d^{5}4s^2$

For, $Mn^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Hg is:-

$[Xe]4f^{14}5d^{10}6s^2$

For, $Hg^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Xe]4f^{14}5d^{10}$

The electronic configuration of the element La is:-

$[Xe]5d^{1}6s^2$

For, $La^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Xe]$

The electronic configuration of the element Fe is:-

$[Ar]3d^{6}4s^2$

For, $Fe^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Ag is:-

$[Kr]4d^{10}5s^1$

For, $Ag^{+}$ , 1 electron is lost, thus the configuration is:-

$[Kr]4d^{10}$

The electronic configuration of the element Co is:-

$[Ar]3d^{7}4s^2$

For, $Co^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{6}$

7 0

3 years ago