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3 years ago

Which of the following is a strong acid? A) HCIO2 B) HF C) HBO3 D) H2SO3 E) H2SO4

Chemistry

1 answer:

ehidna [41]3 years ago

3 0

Answer:

Sulfuric acid

Explanation:

Acid strength (in water) is measured as the equilibria of dissociation of the acid into hydronium and its conjugated base. The greater the Ka value, greater the acid strength.

This equilibrium constant is denominated Ka $Ka=\frac{[H+]}{[HA]}$ where [H+] stands for the molar concentration of hydronium and [HA] stands for the molar concentration of the non-dissociated acid.

The Ka values for these acids are: $1.02x10^{-2}$ for HClO2; $6.4x10^{-4}$ for HF; $5.8x10^{-10}$ for HBO3; $1.2x10^{-2}$ for H2SO3 and $1.x10^{3}$ for H2SO4

So, as the greatest Ka is from H2SO4 it is the strongest acid in this list.

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when copper sulfide is partially roasted in air (reaction with o2), copper sulfite is formed first. subsequently, upon heating,

Lyrx [107]

The question is incomplete, complete question is:

When copper(I) sulfide is partially roasted in air (reaction with oxygen), copper(I) sulfite is formed first. subsequently, upon heating, the copper sulfite thermally decomposes to copper(I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.

Answer:

The balanced chemical equations for these two reactions:

$2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)$

$Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)$

Explanation:

On partial roasting of copper sulfide in an air. The balanced chemical reaction is given as:

$2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)$

On further heating of copper(I) sulfite it get decomposes into copper oxide and sulfur dioxide. The balanced chemical reaction is given as:

$Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)$

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Norma-Jean [14]

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List the following molecules in order from most reducedstate to most oxidizedstate: Formaldehyde (CH2O), Methane (CH4), Carbon D

AleksAgata [21]

Answer:

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

Explanation:

In order to know the oxidation number of C in these compounds, we will use the rule that states that the sum of the oxidation numbers of the atoms is equal to the charge of the compound, in these cases, zero because these are neutral molecules.

H is less electronegative than C, so it acts with the oxidation number +1.
O is more electronegative than C, so it acts with the oxidation number -1. When there is a double bond, it acts with the oxidation number -2.

Formaldehyde (CH₂O)

C + 2.H + O = 0

C + 2.(1) + (-2) = 0

C = 0

Methane (CH₄)

C + 4.H = 0

C + 4.(1) = 0

C = -4

Carbon Dioxide (CO₂)

C + 2.(O-double bond) = 0

C + 2.(-2) = 0

C = +4

Formic Acid (HCOOH)

In this case, we will just consider the H atom linked to the C atom.

C + 1.(H) + 1.(O-double bond) + 1.(O-single bond) = 0

C + 1.(1) + 1.(-2) + 1.(-1) = 0

C = 2

Methanol (CH₃OH)

In this case, we will just consider the H atoms linked to the C atom.

C + 3.(H) + 1.(O-single bond) = 0

C + 3.(1) + 1.(-1) = 0

C = -2

The order from most reduced to most oxidized is:

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

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3 years ago