Answer:
Option A. 1 0n
Explanation:
Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.
Answer: Option (b) is the correct answer.
Explanation:
The elements which have excess or deficiency of electrons will react readily.
Atomic number of Mn is 25 and electronic configuration of
is [Ar]
. This configuration is stable.
Atomic number of Cr is 24 and electronic configuration of
is [Ar]
. This configuration is not stable.
Atomic number of Fe is 26 and electronic configuration of
is [Ar]
. This configuration is stable.
Atomic number of Cu is 29 and electronic configuration of
is [Ar]
. This configuration is not stable.
Atomic number of Al is 13 and electronic configuration of Al is
. This configuration is not stable.
Atomic number of Ba is 56 and electronic configuration of
is [Kr]
. This configuration is stable.
Atomic number of Mg is 12 and electronic configuration of
is
. This configuration is stable.
Atomic number of Sn is 50 and electronic configuration of Sn is [Kr]
. This configuration is stable.
Thus, we can conclude that out of the given options, only Fe and
reactants would lead to a spontaneous reaction as they have incomplete sub-shells. Therefore, in order to gain stability they will readily react.
Explanation:
Below is an attachment containing the solution.
The net ionic equation of borax hydrolysis would be:
<span>Na2B4O7 + 7H2O-----------------> 2 NaOH + 4 H3BO3
</span>
Wherein 2 moles of sodium hydroxide and 4 moles of boric acid are produced by the hydrolysis.
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The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.