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dimaraw [331]
3 years ago
10

Which process will change a substance chemically? A. boiling B. melting C. burning D. evaporating

Chemistry
1 answer:
marshall27 [118]3 years ago
4 0
The answer is B. boiling
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The formation of nitric oxide (N2(g) + O2(g) Imported Asset 2NO(g)) has a Kc value of 4.7 × 10-31. A classmate claims that this
riadik2000 [5.3K]
Answer is: her reasoning is flawed, because <span>Kc is very small, so the concentration of nitric(II) oxide is also very small. </span>

Balanced chemical reaction: N₂(g) + O₂(g) ⇄<span> 2NO(g).
</span>The equilibrium constant<span> (Kc) is a ratio of the concentration of the products (in this reaction nitrogen(II) oxide) to the concentration of the reactants (in this reaction nitrogen and oxygen):
</span>Kc = [NO]² / [N₂] · [O₂].
Kc = 4.7·10⁻³¹.
If we take equilibrium concentration of oxygen and nitrogen to be 1 M:
[N₂] = [O₂] = 1 M.
[NO] = √[N₂] · [O₂] · Kc.
[NO] = 6.855·10⁻¹⁶ M; equilibrium concentration of nitric oxide.
5 0
4 years ago
A prospector pans for gold in a stream.
saveliy_v [14]

Answer:

Mixture

Explanation:

When a prospector scoops up part of the stream bed, the pan contains stream sediments.

3 0
3 years ago
4AI+30 2 &gt; 2AI2O3 What id the chemical reaction?
vesna_86 [32]
This is a combination reaction.  Look at the 2 elements on left and a compound on the right.
7 0
3 years ago
The only way to determine the products of a reaction is to carry out the reaction. All chemical reactions can be classified as o
myrzilka [38]

Answer:

All the statements are correct but "all chemical reactions can be classified as one of the five general types".

Explanation:

Hello,

In this case, I assume you are looking for the wrong statement as long as the following ones are correct and matches with the foundations of chemical reactions:

- The only way to determine the products of a reaction is to carry out the reaction. This is clear, because after the chemical reaction is done, one identifies the products.

- Complete combustion has occurred when all the carbon in the product is in the form of carbon dioxide. This is clear due to a 100% conversion.

- A single reactant is the identifying characteristic of a decomposition reaction. All decomposition reactions have only one reactant which breaks into less complex species.

So the wrong statement is:

- All chemical reactions can be classified as one of five general types. This is wrong because there are four widely known chemical reactions: synthesis, decomposition, simple displacement and double displacement.

Best regards.

8 0
4 years ago
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic
Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<h3>1.33%</h3>

<em />

<em />

<em />

4 0
3 years ago
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