the atomic number of an element x is 14 and its relative atomic mass is 28 write the electronic distribution in an atom of x

Elemental mercury combines with oxygen to make mercury oxide. write a balanced equation

Hg + O2 --> HgO + O

Note that Mercury has a variable valency.

6 0

3 years ago

Now select a longer period, say the fourth period, starting with the element in the 1A group. Again, go from left to right. What

Inessa [10]

Answer:

The pattern of change in the electron configuration increases from left to right across the period.

Explanation:

In the periodic table, elements having the same number of electrons in the outermost shell of their atoms are placed over one another in vertical columns. Each of the vertical columns is known as a group or family. while each of the resulting horizontal rows is known as a period or row.

There are 18 vertical columns and seven periods in the periodic table. Each period begins with an atom having a valence electron and ends with an atom having a complete outer shell structure of an inert gas.

The fourth period of the periodic table (d-block) consists of the transition elements.

The electron distribution into the energy levels of sublevels of atoms can be shown in any of the tree important electronic configuration.

Electronic configuration shows the sequence of filling electrons into the orbitals of the sublevels as guided by some principles.

The maximum number of electrons in the orbitals of sublevels are two for s-sublevel(one orbital) ; six for p-sublevel( three orbital); ten for d-sublevel ( five orbitals) and fourteen for f-sublevel( seven orbitals). This indicates that the maximum number of electrons in an orbital is two

In the Periodic table ,The sublevels with lower energies are filled up before those with higher energies. One important thing about this principle is that the sublevels do not fill up in numerical order. The pattern of filling is as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f etc.

5 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, Δ

victus00 [196]

The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.

We must establish the fact that energy is neither created nor destroyed.

Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.

Total heat absorbed by water, H(water) is;

Q(water) = m C (T2 - T1)

Q(water) = 500 × 4.184 × (100-20)

Q(water) = 167.36 kJ

Consequently, the quantity of heat evolved by the combustion of glycerol is;

Q(glycerol) = 167,360 J = n × ΔH°comb

where, n = no. of moles of glycerol.

167.36 kJ= n × 1654 kJ/mole

n = 167.36/1654

n = 0.1012 moles of glycerol.

Therefore, mass of glycerol combusted, m is;

m = n × Molar mass

m = 0.1012 × 92.09

m = 9.32 g.

the atomic number of an element x is 14 and its relative atomic mass is 28 write the electronic distribution in an atom of x​

the atomic number of an element x is 14 and its relative atomic mass is 28 write the electronic distribution in an atom of x