Tìm Mr Câu 1: 16*2/Mr+16*2=53.3/100 Câu 2: Mr/Mr+1=98.765/100

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1. 1)

   \(\frac{{16.2}}{{{M_R} + 16.2}} = \frac{{53,3}}{{100}} \to 16.2 = \frac{{53,3}}{{100}}.({M_R} + 16.2)\)

   \( \to {M_R} + 32 = 32.\frac{{100}}{{53,3}} = 60\)

   \( \to {M_R} = 28 \to R:Si{\text{ }}(silic)\)

   2)

   \(\frac{{{M_R}}}{{{M_R} + 1}} = \frac{{98,765}}{{100}} \to {M_R} = \frac{{98,765}}{{100}}.({M_R} + 1)\)

   \( \to 0,01235{M_R} = 0,98765\)

   \( \to {M_R} = \frac{{0,98765}}{{0,01235}} = 80 \to R:Br{\text{ (brom)}}\)

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2. Bài giải:

   Câu 1:

   Ta có:$\frac{16.2}{M_R+16.2}=\frac{53,3}{100}$

   $⇔\frac{32}{M_R+32}=\frac{53,3}{100}$

   $⇔\frac{32}{M_R+32}=0,533$

   $⇔32=0,533.(M_R+32)$

   $⇔32=0,533.M_R+17,056$

   $⇔0,533.M_R=14,944$

   $⇔M_R≈28(g/mol)$

   ⇒ $R$ là nguyên tố Silic ($Si$)

   Câu 2:

   Ta có: $\frac{M_R}{M_R+1}=\frac{98,765}{100}$ 

   $⇔\frac{M_R}{M_R+1}=0,98765$ 

   $⇔M_R=0,98765.(M_R+1)$

   $⇔M_R=0,98765.M_R+0,98765$

   $⇔0,012345.M_R=0,98765$

   $⇔M_R≈80(g/mol)$

   ⇒ $R$ là nguyên tố Brôm ($Br$)

    

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