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lubasha [3.4K]
3 years ago
15

Which of the following shows an accurate combustion reaction?

Chemistry
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

  • <em>Cu + O₂  → CuO₂</em>

Explanation:

A <em>combustion reaction</em> is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

  • <em>Cu + O₂ → CuO₂</em>

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

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2 years ago
In a sulphuric acid (h2so4) - sodium hydroxide (naoh) acid-base titration, 17.3 ml of 0.126 m naoh is needed to neutralize 25 ml
katen-ka-za [31]
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted  - 0.126 mol/L  x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄ 
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄ 
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol 
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4 0
3 years ago
37.9 grams of an unknown substance undergoes a temperature increase of
jenyasd209 [6]

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

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3 years ago
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Answer:

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In Balancing Equations, why do I put a 3 before reactant H?
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