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Ymorist [56]
3 years ago
14

How is burning magnesium different than burning methane

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
6 0

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2  O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

brainly.com/question/13824679

#learnwithBrainly

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A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and<br> 32 C.
lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁  = 928kpa

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Unknown:

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Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

           \frac{P_{1} V_{1} }{T_{1} }   = \frac{P_{2} V_{2} }{T_{2} }

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

 take the units to the appropriate ones;

             kpa to atm,  °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁  = 928kpa gives 9.16atm

T₂  = 32°C gives 273 + 32  = 305K

T₁  = 129°C gives 129 + 273  = 402K

Input the values in the equation and solve for V₂;

        \frac{9.16  x 569}{402}   = \frac{3.15 x V_{2} }{305}

       V₂   = 1255.4L

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Explanation:

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