Answer:
3.33 tanques de O₂
Explanation:
Basados en la reacción:
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)
<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>
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La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:
9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>
Si un tanque contiene 7x10³ L de O₂ serán necesarios:
23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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It depends on the pH if the base. but normally light colors are for bases example blue green etc
Answer:
C. 4.00 K
Explanation:
We can solve this using Charles's Law of the ideal gas. The law describes that when the pressure is constant, the volume will be directly proportional to the temperature. Note that the temperature here should only use the Kelvin unit. Before compressed, the volume of the gas is 50ml(V1) and the temperature is 20K (T1). After compressed the volume becomes 10ml(V2). The calculation will be:
V1 / T1= V2 / T2
50ml / 20K = 10ml / T2
T2= 10ml/ 50ml * 20K
T2= 4K