Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
Answer: decrease the temperature of the reaction
Explanation:
CH3COOH + CH3CH2OH ⇌ CH3COOCH2CH3 + H2O
The formation of an ester is an exothermic reaction. Therefore, a decrease in the temperature of the reaction will favour the production of ester.
The question is incomplete. Complete question is attached below:
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Correct Answer:
Option A i.e FlammabilityReason:
Hydrochloric acid is a
strong acid. It has a high value of dissociation constant. Following equilibrium is established with HCl is dissolved in water
HCl + H2O ↔ Cl- + H3O+
Above, reaction is highly exothermic in nature. So care must be taken while preparing HCl solution of desired concentration.
HCl must be added in beaker containing water. Reversed, i.e. adding water in HCl, must never be done. This is because when acid is poured into water, it flows down the flask and mixes much efficiently and no boiling occurs. On other hand, <span> when water is added to acid, the very first drop of water will react with acid. The heat generated during this event might be high enough to boil the water instantly, which could spray acid out of the container.
The physical contact with concentration HCl will result in skin burn. As well, inhaling</span><span> HCl vapors can cause burns to body tissues, lung damage and even death in extreme cases. Also, since HCl is a strong acid, it would undergoes reaction almost spontaneously, which would inturn favour substitution reaction. However,
HCl is not a flammable gas. Hence, it is of no concern while performing substitution reaction. </span>
