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Korolek [52]
3 years ago
15

A crystal is a solid (or liquid) in which the constituent atoms, molecules, or ions are packed to have a ________________ as wel

l as a short range order
Chemistry
2 answers:
yan [13]3 years ago
6 0

Answer:

long range order

Explanation:

A crystal consists of atoms, ions or molecules having both short range and long range order. The atoms, ions or molecules are arranged in a regular pattern throughout the lattice both at immediate vicinities and across the entire crystal structure.

This order accounts for the definite shape and unique properties of crystals which include their sharp melting and boiling points which distinguishes them from amorphous substances.

Leni [432]3 years ago
4 0

Answer:

long range order.

Explanation:

Crystals are atoms or molecules that are arranged in a manner which makes them repeats in a definite way(s). Crystals can form shape or structure such as hexagonal shape, monoclinic structure, cubic shape, tetragonal shape, orthorhombic shape and so on.

Long range order packing of crystals simply describe one of the properties of crystalline solids which is the regular repetition of a of certain arrangement pattern periodically.

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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

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Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

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As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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