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balu736 [363]
3 years ago
13

A mixture in which one substance is evenly mixed with another substance is called a..

Chemistry
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

C solution

Explanation:

i hope that i got it right im not very good with questions

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)

Substance      ΔG°f (kJ/mol)

M₃O₄                 -9.50

M(s)                       0

O₂(g)                     0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

PCl_3(l)\rightarrow PCl_3(g)

First we have to calculate the Gibbs energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]

where,

\Delta G^o = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]

\Delta G^o=9.50kJ/mol

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs free energy  = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K  = equilibrium constant = ?

9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)

K=0.0216

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

K=(P_{O_2})^2

0.0216=(P_{O_2})^2

P_{O_2}=0.147atm

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0
4 years ago
There are two main characters of matter:
Oksana_A [137]
Mass and weight are the 2 main characters of matter
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Which subatomic particles takes up most of the space of an atom
Vadim26 [7]
The neutrons take up most of the space of an atom
7 0
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ClO(4)− Express your answer as a chemical formula.
Nimfa-mama [501]

ClO(4)− is regarded as perchlorate ion and are produced commercially in most situations as salts via industries and in the laboratory.

<h3>What is Perchlorate ion?</h3>

This ion is referred to as a monovalent inorganic anion and is obtained by deprotonation of perchloric acid. It is composed of chlorine and oxygen atoms in the ratio 1 to 4 respectively.

This has  32 valence electrons available in the Lewis structure and is used in the commercial production of solid rocket fuel.This ion has a molar mass of 99.451 g mol−1 and is used in different processes such as an oxidizer and to control static electricity during the process of food preservation in industries.

Therefore ClO₄− is also regarded as perchlorate ion and is the most appropriate choice.

Read more about Perchlorate ion here brainly.com/question/16895150

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3 0
2 years ago
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
Sladkaya [172]

Answer:

Part A:

Charge is P^{3-}

Configuration is 1s^2 2s^22p^63s^23p^6

Part B:

Charge is Mg^{2+}

Configuration is 1s^2 2s^22p^6

Part C:

Charge is Se^{2-}

Configuration is 1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6

Explanation:

Monatomic ions:

These ions consist of only one atom. If they have more than one atom then they are poly atomic ions.

Examples of Mono Atomic ions: Na^+, Cl^-, Ca^2^+

Part A:

For P:

Phosphorous (P) has 15 electrons so it require 3 more electrons to stabilize itself.

Charge is P^{3-}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^63s^23p^6

Part B:

For Mg:

Magnesium (Mg) has 12 electrons so it requires 2 electrons to lose to achieve stable configuration.

Charge is Mg^{2+}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^6

Part C:

For Se:

Selenium (Se) has 34 electrons and requires two electrons to be stable.

Charge is Se^{2-}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6

8 0
3 years ago
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