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
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Answer:
<u>It increases by a factor of four</u>
Explanation:
Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.
pV = K.......(1)
pV = constant
Charles law : The volume of the gas is directly proportional to temperature at constant pressure.
V = KT
or V/T = K = constant ....(2)
Applying equation (1) and (2)
According to question ,
T2 = 4 (T1)
V2 = V1
Put the value of T2 and V2 , The P2 can be calculated,
V1 and V1 cancel each other
T1 and T1 cancel each other
We get,
or
P2 = 4 P1
So pressure increased by the factor of four
Answer:
1.0975 atm.
Explanation:
<em>According to Boyle’s Law:</em> "
At constant temperature , the volume of a given quantity of a gas varies inversely with its pressure".
P α 1/V.
<em>∴ P₁V₁ = P₂V₂.</em>
P₁ = 4.39 atm, V₁ = 0.5 L.
P₂ = ??? atm, V₂ = 2.0 L.
<em>∴ P₂ = P₁V₁/V₂</em> = (4.39 atm)(0.5 L) / (2.0 L) = <em>1.0975 atm.</em>
Hey there!:
H is always +1 so the H's have a +3 charge.
O is always -2 so the O's have a -8 charge .
Now, suppose oxidation state for P = X , then :
+3 + X + (-8) = 0 (because of neutral molecule)
x = 8 - 3
x = + 5
So, X = +5 oxidation state.
Answer C
Hope that helps!
Answer:Um... I think 5000 i am not really sure
Explanation: I Dont Really Know
