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FromTheMoon [43]
2 years ago
13

A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white crystalline substance B an

d a gas C. The gas ignites readily when combined with hydrogen gas and a liquid uncolored substance resulted. Based on these observations, can we determine whether solids A and B and gas C, are elements or compounds?
Chemistry
1 answer:
attashe74 [19]2 years ago
3 0

Answer:

A and B as well C are the compounds.

Explanation:

  • A seems to be a compound, very least, it appears to contain carbon and oxygen.
  • C seems to be a compound, containing respectively oxygen as well as carbon.
  • B isn't always described by that of the information provided, it is therefore presumably a compound because there are few components as white solid particles.

So that the above is the correct answer.

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We know this because of the discovery of fossils and bones. We can find out roughly when they existed because of carbon dating.
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What happens when a star dies
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They become a supernova

Explanation:

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What are some types of chemical weathering?​
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Carbonation. When you think of carbonation, think carbon! ...

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Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu
borishaifa [10]

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

6 0
3 years ago
K for enolization of 2,4-cyclohexadienone is about 1013. Explain why the enol is so much more stable than the keto tautomer.
torisob [31]

Answer:

Because the value of K is huge.

Explanation:

The tautomer is a kind of isomer in which exist an equilibrium between a ketone and an enol, or between an aldehyde and an enol. So, in the enolization, the ketone is the reactant and the enol is the product.

The equilibrium reaction can be characterized by an equilibrium constant, which is the ratio of the concentration of the products by the concentration of the reactants.

Because the constant K is extremely large (10¹³) we can conclude that the concentration of the product will be greater than the concentration of the reactant, in the equilibrium. It means that the concentration of the enol will be greater.

So, the ketone is unstable and forms in a great amount the more stable product, the enol.

4 0
3 years ago
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