How does the rate of reaction proceed with time

For the reaction: 2 A (g) + B (s)= 2 C(s) + D (g)

jolli1 [7]

Answer:

The value of the equilibrium constant = 5.213

Explanation:

Here $K_p$ (equilibrium constant) is referred to as the partial pressure of product divided by the partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation .

As such only gas appear in $K_p$ expression as solids takes a value of 1;

SO ; in the given equation from the question:

2 A (g) + B (s) ----> 2 C(s) + D (g)

$K_p = \dfrac{[D]}{[A]^2}$

$K_p = \dfrac{2.71}{0.721^2}$

$K_p = 5.213$

The value of the equilibrium constant = 5.213

8 0

3 years ago

The elements cu, o, la, y, ba, tl, and bi are all found in high-temperature ceramic superconductors. write the expected electron

timama [110]

Part 1)
Cu- [Ar] 3d¹⁰4s¹ atomic number: 29

O- [He] 2s2 2p4 atomic number:8

La- [Xe] 5d¹ 6s² atomic number:57

Y- [Kr] 4d¹5s² atomic number:39

Ba- [Xe] 6s² atomic number:56

Tl- [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹ atomic number:81

Bi- [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p³ atomic number:83

Part 2)
You are able to this by consulting the periodic table and following this steps:
-Find your atom's atomic number;
-Determine the charge of the atom (these were all uncharged)
-Memorize the order of orbitals (s, d, p, d.. and how many electrons they can fit)
-Fill in the orbitals according to the number of electrons in the atom
- for long electron configurations, abbreviate with the noble gases

4 0

3 years ago

Ions that have the same electron configuration are called

Katena32 [7]

Such ions are said to be isoelectronic

3 0

3 years ago

Which symbol equation correctly describes the reaction: Wood (mostly glucose) burns in the presence of oxygen to form water and

SCORPION-xisa [38]

Answer:

Explanation:

Glucose = C6H12O6

oxygen = O2

water = H2O

carbon dioxide = CO2

balanced equation :

C6H12O6 (s) + 6 O2 (g) --------------> 6 CO2 (g) + 6 H2O (g)

7 0

3 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

3 years ago

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