Which will be chemically more reactive: element $A$ with atomic number $17$ or element $B$ with atomic number $16$? Give a reason as well.
(A) $(i)$ Element $A$ (Chlorine) has an atomic number of $17$,and its electronic configuration is $(2, 8, 7)$. It needs to gain $1$ electron to complete its octet.
$(ii)$ Element $B$ (Sulfur) has an atomic number of $16$,and its electronic configuration is $(2, 8, 6)$. It needs to gain $2$ electrons to complete its octet.
$(iii)$ Reactivity in non-metals is determined by the ease of gaining electrons. Since element $A$ requires only $1$ electron to achieve a stable noble gas configuration,it is more chemically reactive than element $B$,which requires $2$ electrons.
$(ii)$ Element $B$ (Sulfur) has an atomic number of $16$,and its electronic configuration is $(2, 8, 6)$. It needs to gain $2$ electrons to complete its octet.
$(iii)$ Reactivity in non-metals is determined by the ease of gaining electrons. Since element $A$ requires only $1$ electron to achieve a stable noble gas configuration,it is more chemically reactive than element $B$,which requires $2$ electrons.
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Express each of the following in kilograms:
$(a)$ $5.84 \times 10^{-3} \,mg$
$(b)$ $58.34 \,g$
$(c)$ $0.584 \,g$
$(d)$ $5.873 \times 10^{-21} \,g$
$(a)$ $5.84 \times 10^{-3} \,mg$
$(b)$ $58.34 \,g$
$(c)$ $0.584 \,g$
$(d)$ $5.873 \times 10^{-21} \,g$
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$(i)$ In a balanced chemical equation,the sum of the masses of reactants and products remains unchanged.
$(ii)$ $A$ group of atoms carrying a fixed charge on them.
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$(ii)$ Ammonium carbonate
$(b)$ Write one word for the following:
$(i)$ In a balanced chemical equation,the sum of the masses of reactants and products remains unchanged.
$(ii)$ $A$ group of atoms carrying a fixed charge on them.
$(c)$ Write chemical formulae of the following compounds:
$(i)$ Sodium phosphate
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View SolutionFill in the missing data in the Table.
Species property $H_2O$ $CO_2$ $Na$ atom $MgCl_2$ No. of moles $2$ - - $0.5$ No. of particles - $3.011 \times 10^{23}$ - - Mass $36 \, g$ - $115 \, g$ -
| Species property | $H_2O$ | $CO_2$ | $Na$ atom | $MgCl_2$ |
| No. of moles | $2$ | - | - | $0.5$ |
| No. of particles | - | $3.011 \times 10^{23}$ | - | - |
| Mass | $36 \, g$ | - | $115 \, g$ | - |
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$(i)$ $100 \, g$ of $N_2$
$(ii)$ $100 \, g$ of $NH_3$
$(i)$ $100 \, g$ of $N_2$
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View SolutionAs per the law of definite proportion,carbon and oxygen combine in the ratio of $3:8$. Compute the mass of oxygen gas that would be required to react completely with $15\,g$ of carbon.
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