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3 years ago

Which of the following is an essential condition for a combustion reaction ?

Chemistry

2 answers:

Ugo [173]3 years ago

8 0

B is the answer to your question

WARRIOR [948]3 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

A combustion reaction is defined as the reaction in which a substance chemically reacts in the presence of oxygen.

A combustion reaction is always exothermic in nature as heat is released. Also, carbon dioxide and water will be produced in a combustion reaction.

For example, $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Therefore, we can conclude that an essential condition for a combustion reaction is that oxygen is a reactant.

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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago

Please help! (The question is in the picture above)

goldfiish [28.3K]

(i) The pH of HCl is between 0 and 2.
(ii) the presence of Carbon Dioxide turns lime water milky or sees a white precipitate form on the surface of the lime water. Lime water is CaOH and when it reacts with Carbon Dioxide it forms a white solid (precipitate) Calcium Carbonate.
$Ca(OH)_{2} _{(l)} + CO_{2} _{(g)} -----\ \textgreater \ CaCO_{3} _{(s)} + H_{2}O_{(l)}$

7 0

3 years ago

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid

bogdanovich [222]

Explanation:

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). They will react according to the following equation.

HBr + NaOH ---> NaBr + H₂O

0.81 g of HBr are mixed with 0.568 g of NaOH. We have to find the mass of NaBr that can be produced. To do that we have to find which of the reactants is limiting the reaction. First, we will convert their grams into moles using their molar masses.

molar mass of HBr = 80.91 g/mol

molar mass of NaOH = 40.00 g/mol

mass of HBr = 0.81 g

mass of NaOH = 0.568 g

moles of HBr = 0.81 g * 1 mol/(80.91 g)

moles of HBr = 0.0100 moles

moles of NaOH = 0.568 g * 1 mol/(40.00 g)

moles of NaOH = 0.0142 moles

HBr + NaOH ---> NaBr + H₂O

Now if we take a quick look at the coefficients of the reaction we will see that 1 mol of HBr will react with 1 mol of NaOH since both coefficients are 1. Then their molar ratio is 1 : 1. That also means that 0.0100 moles of HBr will only react with 0.0100 moles of NaOH, and we have mixed 0.0142 moles of it. So, NaOH is in excess and HBr is the limiting reagent.

1 mol of HBr : 1 mol of NaOH molar ratio

moles of NaOH = 0.0100 moles of HBr * 1 mol of NaOH/(1 mol of HBr)

moles of NaOH = 0.0100 moles < 0.0142 moles ----> NaOH is in excess

And now that we know that HBr is the limiting reagent we can find the number of moles of NaBr that will be produced by 0.0100 moles of HBr. And finally convert those moles into grams using the molar mass.

1 mol of HBr : 1 mol of NaBr molar ratio

moles of NaBr = 0.0100 moles of HBr * 1 mol of NaBr/(1 mol of HBr)

moles of NaBr = 0.0100 moles

molar mass of NaBr = 102.89 g/mol

mass of NaBr = 0.0100 moles * 102.89 g/mol

mass of NaBr = 1.0289 g

mass of NaBr = 1.0 g

Answer: the maximum mass of sodium bromide that could be produced is 1.0 g.

7 0

10 months ago

3 why does soldium produce blue colour when dissolve in ammonia?

bogdanovich [222]

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6 0

2 years ago

Fig. 6.1 shows the structural formula of an organic compound.

Tatiana [17]

Answer:

A dono wat is da anser sori man hehe

6 0

2 years ago

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