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Brut [27]
2 years ago
15

1)The metals of group 2 from top to bottom are: Be, Mg, Ca, Sr,Ba. Which if these metal will form ions most readily and why?

Chemistry
1 answer:
Pepsi [2]2 years ago
7 0

1) <em>Most ionizable atom </em>

Atomic size increases as you go down a Group (see image 1). We are adding electrons to increasingly larger shells.

The <em>valence electrons are further from the attraction of the nucleus</em>, so they are less tightly held.

Thus, Ba will form ions most readily.

2) <em>Ionization energy </em>

As an atom gets larger, the electrons are less strongly attracted to the nucleus, so it takes less energy to remove the valence electrons.

<em>Ionization energies decrease from top to bottom</em> of a Group in the Periodic Table.

Thus, in Group 2, Be has the highest ionization energy, and it decreases as you go down the Group. Ba has the lowest ionization energy.

3) <em>Ionization energy exceptions </em>

In general, ionization energies increase from left to right in a period.

However, the ionization energies of B and O are smaller than that of their immediate predecessors (see image 2).

The ionization energy of B is less than that of Be because B has a filled 2s orbital, which increases the shielding of the 2p electron and reduces the ionization energy.

The ionization energy of O is less than that of N because we are adding an extra electron to a half-filled 2p subshell.

The extra electron must go into an orbital that already contains another electron. The increased repulsion raises the energy level.

It becomes easier to remove an electron, so the ionization energy decreases.

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Nikolay [14]

Explanation:

Below is an attachment containing the solution.

7 0
3 years ago
Can you pls tell me the net ironic equation of H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)
almond37 [142]

Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

5 0
2 years ago
Which substance is the oxidizing agent in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which substance
aleksandr82 [10.1K]

Answer:

HNO3 is the oxidizing agent

Explanation:

Step 1:

The oxidizing agent is responsible to oxidize another.  but  itself undergoes  a reduction  .

Fe2S3:

Fe has an oxidation number of +3

S has an oxidation number of -2

HNO3:

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

H has an oxidation number of +1

N has an oxidation number of +5

Fe(NO3)3

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

Fe has an oxidation number of +3

Since NO3 has an oxidation number of -1; N has an oxidation number of +5

S alone has an oxidation number of 0

NO2:

O has an oxidation number of -2 (we have 2 times O, this makes -4)

N has an oxidation number of +4

Fe doesn't change from oxidation number. It stays +3

N goes from +5 to +4 → this is a reduction

S goes from -2 to 0 → this is an oxidation

The reducing agent is the compound that contributes the oxidized species (S).

The oxidizing agent contributes the reducing species (N)

The answer is HNO3

5 0
2 years ago
A mixture of carbon dioxide and helium gases is maintained in a 7.91 L flask at a pressure of 1.42 atm and a temperature of 33 °
Crank

Answer:

The gas mixture contains 1.038 grams of helium

Explanation:

<u>Step 1:</u> Data given

Volume of the flask = 7.91 L

Total Pressure = 1.42 atm

Temperature = 33 °C

Mass of CO2 = 8.25 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of He = 4 g/mol

<u>Step 2</u>: Calculate total number moles of gas

p*V = n*R*T

⇒ p = the pressure = 1.42 atm

⇒ V = the volume = 7.91 L

⇒ n= the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L* atm/K*mol

⇒ T = the temperature = 33 °C = 306 Kelvin

n = (p*V)/(R*T)

n = (1.42*7.91)/(0.08206 * 306)

n = 0.447 moles

<u>Step 3</u>: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 8.25 grams / 44.01 g/mol

Moles CO2 = 0.1875 moles

<u>Step 4:</u> Calculate moles of Helium

Moles Helium = total moles of gas - moles of CO2

Moles Helium = 0.447 - 0.1875 = 0.2595 moles of helium

<u>Step 5: </u> Calculate mass of helium

Mass of helium = moles of helium * molar mass of helium

Mass of helium = 0.2595 moles * 4 g/mol

Mass of helium = 1.038 grams

The gas mixture contains 1.038 grams of helium

8 0
2 years ago
What volume of carbon dioxide gas can be collected from
alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

  • CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we <u>determine which reactant is limiting</u>:

  • Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
  • Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

  • 0.05 mol HCl * \frac{1molCO_{2}}{1molHCl} = 0.05 mol CO₂

Finally we use PV=nRT to <u>calculate the volume</u>:

  • P = 1 atm
  • n = 0.05 mol
  • T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

  • V = 1.22 L
7 0
3 years ago
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