Question
Convert the following hexadecimal numbers into their decimal equivalents:(a) $7 \mathrm{~A}_{16}$(b) $3 \mathrm{~F}_{16}$(a) $7 \mathrm{~A}_{16}=7 \times 16^{1}+\mathrm{A} \times 16^{0}=7 \times 16+10 \times 1$ $=112+10=122$Thus $\mathbf{7 A}_{16}=\mathbf{1 2 2}_{10}$(b) $3 \mathrm{~F}_{16}=3 \times 16^{1}+\mathrm{F} \times 16^{0}=3 \times 16+15 \times 1$ $=48+15=63$Thus $3 \mathrm{~F}_{16}=63_{10}$
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The rightmost digit represents $16^0$, the next digit to the left represents $16^1$, and so on. Show more…
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Convert the following hexadecimal numbers into their decimal equivalents: (a) $7 \mathrm{~A}_{16}$ (b) $3 \mathrm{~F}_{16}$ (a) $7 \mathrm{~A}_{16}=7 \times 16^{1}+\mathrm{A} \times 16^{0}=7 \times 16+10 \times 1$ $=112+10=122$ Thus $\mathbf{7 A}_{16}=\mathbf{1 2 2}_{10}$ (b) $3 \mathrm{~F}_{16}=3 \times 16^{1}+\mathrm{F} \times 16^{0}=3 \times 16+15 \times 1$ $=48+15=63$ Thus $3 \mathrm{~F}_{16}=63_{10}$
Convert the following hexadecimal numbers into their decimal equivalents: (a) $\mathrm{C} 9_{16}$ (b) $\mathrm{BD}_{16}$ (a) $\mathrm{C} 9_{16}=\mathrm{C} \times 16^{1}+9 \times 16^{0}=12 \times 16+9 \times 1$ $$ =192+9=201 $$ Thus C9_{16 } = 2 0 1 _ { 1 0 } (b) $\mathrm{BD}_{16}=\mathrm{B} \times 16^{1}+\mathrm{D} \times 16^{0}$ $=11 \times 16+13 \times 1=176+13=189$ Thus $\mathbf{B D}_{16}=\mathbf{1 8 9}_{10}$
Convert the following hexadecimal numbers into their binary equivalents: (a) $3 \mathrm{~F}_{16}$ (b) $\mathrm{A} 6_{16}$ (a) Spacing out hexadecimal digits gives: $\overbrace{0011}^{3} \overbrace{1111}^{\mathrm{F}}$ and converting each into binary gives as above, from Table $10.2$. Thus, $3 \mathrm{~F}_{16}=\mathbf{1 1 1 1 1 1}_{2}$ (b) Spacing out hexadecimal digits gives: $\overbrace{1010}^{\mathrm{A}} \overbrace{01,10}^{6}$ and converting each into binary gives as above, from Table $10.2$. Thus, $\mathbf{A 6}_{16}=\mathbf{1 0 1 0 0 1 1 0}_{2}$
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