I believe the major periodic trends include; electronegativity, ionization energy, electron affinity, atomic radius, melting point, and the metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's property.

3 0

3 years ago

What is the purpose of a Lewis structure?

docker41 [41]

Answer:

Explanation:

in a lewis structure, it shows the connectivity of a molecule as well as how many and where the valence electrons are located.

6 0

2 years ago

For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t

Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) $2.136197(10^{25})$ Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = $6.02.(10^{23})$ elemental units

For example :

1 mol of oxygen atoms = $6.02.(10^{23})$ oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

$molarmass_{(Al)}=26.982\frac{g}{mol}$

$molarmass_{(S)}=32.065 \frac{g}{mol}$

$molarmass_{(O)}=15.999\frac{g}{mol}$

If we sum all the molar masses = $(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}$

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

$2.663 kg=2663g$

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

$x=\frac{2663}{75.046}mol=35.485 mol$

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = $6.02(10^{23})$

35.485 moles of Al have $(35.485)(6.02)(10^{23})=2.136197(10^{25})$ Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

$x=\frac{(2663).(15.999)}{75.046} g$

x = 567.723 g of O

6 0

3 years ago

Hydrogen can be extracted from natural gas according to the following equilibrium.

antiseptic1488 [7]

Answer:

2.16x10⁻²

Explanation:

First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:

nCH4 = 24.0/16 = 1.5 moles

nCO2 = 88.0/44 = 2 moles

The concentration is the number of moles divded by the volume, thus:

[CH4] = 1.5/1 = 1.50 M

[CO2] = 2/1 = 2.00 M

For the equilibrium reaction, let's do an equilibrium chart:

CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)

1.50 2.00 0 0 Initial

-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)

1.50-x 2.00-x 2x 2x Equilibrium

As sateted in problem, [CH4] = 2.70*[CO]

1.50 - x = 2.70*2x

1.50 - x = 5.4x

6.4x = 1.50

x = 0.2344

Thus, at equilibrium:

[CH4] = 1.50 - 0.2344 = 1.2656 M

[CO2] = 2.00 - 0.2344 = 1.7656 M

[CO] = 2*0.2344 = 0.4688 M

[H2] = 2*0.2344 = 0.4688 M

The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.

K = ([CO]²*[H2]²)/([CH4]*[CO2])

K = (0.4688²*0.4688²)/(1.2656*1.7656)

K = 0.0483/2.2345

K = 2.16x10⁻²

7 0

3 years ago