All categories

3 years ago

What effect does heating a solid to a hightemperature have have on its crystal lattice ??

Chemistry

2 answers:

allsm [11]3 years ago

8 0

Answer:

The lattice breaks apart and the ions flow freely.

Explanation:

kotegsom [21]3 years ago

6 0

Arrangement of atoms or group of atoms in a three dimensional ordered pattern in a crystal is said to be a crystalline lattice. They are arranged in a specific pattern with high symmetry. The heating of the crystal at high temperature will result in the increase of vibrational kinetic energy of the atoms in the crystal and this will result in the breaking of lattice apart and due to the breaking of lattice apart the ions will flow freely.

Thus, the heating of a solid at high temperature will lead to the lattice breaks apart and ions will flow freely.

You might be interested in

What is the % yield when 12g of Mg react with HCl to produce 0.2g of H2?

goblinko [34]

Answer:

The yield is 20 %.

Explanation:

Mg + 2 HCl → MgCl2 + H2

(12 g Mg) / (24.30506 g Mg/mol) x (1 mol H2 / 1 mol Mg) x (2.015894 g H2/mol) = 0.9953 g H2.

(0.2 g) / (0.9953 g) = 0.20 = 20% yield

5 0

2 years ago

What is the formula for barium nitrate ?

brilliants [131]

Answer:

Ba(NO3)2

Explanation:

6 0

3 years ago

Common units of electrical force are: A coulombs B Newtons C meters

dybincka [34]

b is the correct answer

4 0

3 years ago

Read 2 more answers

write the balanced equation for the reaction that occurs when butane undergoes combustionWhat is the coefficient of oxygen in th

love history [14]

Answer:

2C4H10 + 13O2 ----> 4CO2 + 10H2O

The coefficient of oxygen in the balanced equation is 13

5 0

3 years ago

The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

3 0

2 years ago