The crystal structure of an ionic compound depends on the

Chemistry

1 answer:

Murljashka [212]3 years ago

4 0

Answer:

The ion's sizes and their radius ratio.

Explanation:

Ionic compounds take the form of three dimentional arrays by alternating anions and cations in order to minimize the potential energy of the system by maximizing the attractive force between opposite charges.

<em>The resultant geometric structures are known as crystal lattices and their arrangements depend on the ion's sizes and, since they are bound though electrostatic attraction, their radius ratio.</em>

I hope you find this information useful and interesting! Good luck!

You might be interested in

Container A (with volume 1.23 L) contains a gas under 3.24 atm of pressure. Container B (with volume

kolbaska11 [484]

Now that a lot of pressure

8 0

3 years ago

Which one of the following questions about animals called affairs, pictured above, is the scientific question?

tino4ka555 [31]

The answer is B...

because any of the others could have an opinion based answer but B will always be the same its a fact not opinion.

3 0

3 years ago

How do you know that a bond between magnesium and chlorine will be ionic?

OverLord2011 [107]

An ionic bond is a bond between a metal and non-metal
Considering that magnesium is a metal and chlorine is a non-metal an ionic bond will form.

8 0

3 years ago

Hydrogen and nitrogen react to produce ammonia gas as shown in the following chemical

DerKrebs [107]

Explanation:

$31.5 \: g \: aommnia \: \times \: \frac{1 \: mol \: ammonia}{8 \: g \: ammonia \: } \: \times \frac{3 \: mol \: hydrogen}{2 \: mol \: ammonia} \: \times \frac{1 \: g \: hydrogen}{1 \: mol \: hydrogen} = \: 5.90625 \: g \: hydrogen \: must \: react \:$

5 0

2 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

8 0

3 years ago