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abruzzese [7]
4 years ago
6

A chemical reaction was used to separate a sample into its components. What conclusion about the sample can be formed based on t

he technique used? (4 points)
a It was a pure substance.
b It was a compound.
c It was a heterogeneous mixture.
d It was a homogeneous mixture
Chemistry
2 answers:
Anni [7]4 years ago
7 0

Answer:

b It was a compound.

Explanation:

<em>A chemical reaction was used to separate a sample into its components. What conclusion about the sample can be formed based on the technique used? </em>

<em>a It was a pure substance.</em>  NO. If the pure substance is a compound, this option could be. However, if the pure substance is a simple substance, it could not be separated into its components.

<em>b It was a compound. </em> YES. Compounds can be separated into pure substances using chemical reactions. For instance, H₂O → H₂ + 1/2 O₂

<em>c It was a heterogeneous mixture.</em>  NO. Heterogeneous mixtures can be separated into homogeneous mixtures using physical processes such as sieving, filtration, etc.

<em>d It was a homogeneous mixture.</em> NO. Homogeneous mixtures can be separated into pure substances using physical processes such as distillation, chromatography, etc.

frez [133]4 years ago
3 0

It is A I took a test just like this yesterday.

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While bond energies and bond enthalpies can be used to estimate the heat of reaction (enthalpy change of a reaction), H, the heat of neutralization is the heat released when 1 mole of water is generated by the reaction of an acid and a base (reaction).

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olga55 [171]

Explanation:

Since, the given reaction is as follows.

       2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)

Initial:    36.1 atm                 0          0

Change:    2x                      x           x

Equilibrium: (36.1 - 2x)       x            x

Now, expression for K_{p} of this reaction is as follows.

            K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of O_{2} at equilibrium will be calculated as follows.

              K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

        2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}

                 x = 18.1 atm

Thus, we can conclude that partial pressure of O_{2} at equilibrium is 18.1 atm.

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