Question

calcu M pentru următoarele substanțe :trioxid de fosfor, dioxid de sulf, oxid de calciu, oxid de sodiu, hidroxid de magneziu, hidroxid de potasiu, acid carbonic, acid azotos, acid fosforic, hidrogen sulfurat, azot (N2), sulfat de cupru, carbonat de sodiu, sulfat de aluminiu. Va roggggggg​

Answer 1

Cerință:

Calculați M pentru următoarele substanțe:

• trioxid de fosfor - P₂O₃;
• dioxid de sulf - SO₂;
• oxid de calciu - CaO;
• oxid de sodiu - Na₂O;
• hidroxid de magneziu - Mg(OH)₂;
• hidroxid de potasiu - KOH;
• acid carbonic - H₂CO₃;
• acid azotos - HNO₂;
• acid fosforic - H₃PO₄;
• hidrogen sulfurat - H₂S;
• azot - N₂;
• sulfat de cupru - CuSO₄;
• carbonat de sodiu - Na₂CO₃;
• sulfat de aluminiu - Al₂(SO₄)₃;

Răspuns:

$\LARGE\color{red}\boxed{\bf \mathcal M_{P_2O_3}=110\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{SO_2}=64\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{CaO}=56\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{Na_2O}=62\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{Mg(OH)_2}=58\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{KOH}=56\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{H_2CO_3}=62\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{HNO_2}=47\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{H_3PO_4}=98\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{H_2S}=34\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{N_2}=28\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{CuSO_4 }=160 \dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{Na_2CO_3 }=106\dfrac{g}{mol}}$

$\LARGE\color{red}\boxed{\bf \mathcal M_{Al_2(SO_4)_3}=342\dfrac{g}{mol}}$

Explicație:

Masa molară se calculează însumând masele atomice relative ale elementelor componente substanței. Se notează cu M și are unitatea de măsură g/mol.

Calcule:

$\bf \mathcal M_{P_2O_3}=2\cdot A_{P}+3 \cdot A_{O}=2\cdot 31+3\cdot 16=62+48=110\dfrac{g}{mol}$

$\bf \mathcal M_{SO_2}=A_{S}+2\cdot A_{O}=32+2\cdot16=32+32=64 \dfrac{g}{mol}$

$\bf \mathcal M_{CaO}=A_{Ca}+A_{O}=40+16=56 \dfrac{g}{mol}$

$\bf \mathcal M_{Na_2O}=2\cdot A_{Na}+A_{O}=2\cdot23+16=46+16=62 \dfrac{g}{mol}$

$\bf \mathcal M_{Mg(OH)_2}=A_{Mg}+2\cdot A_{O}+2\cdot A_{H}=24+2\cdot16+2\cdot1=24+32+2=58 \dfrac{g}{mol}$

$\bf \mathcal M_{KOH}=A_{K}+A_{O}+A_{H}=39+16+1=56 \dfrac{g}{mol}$

$\bf \mathcal M_{H_2CO_3}=2\cdot A_{H}+A_{C}+3\cdot A_{O}==2\cdot1+12+3\cdot16=2+12+48=62 \dfrac{g}{mol}$

$\bf \mathcal M_{HNO_2}=A_{H}+A_{N}+2\cdot A_{O}=1+14+2\cdot16=15+32=47 \dfrac{g}{mol}$

$\bf\mathcal M_{H_3PO_4}=3\cdot A_{H}+A_{P}+4\cdot A_{O}=3\cdot1 +31+4\cdot16=3+31+64=98\dfrac{g}{mol}$

$\bf\mathcal M_{H_2S}=2\cdot A_{H}+A_{S}=2\cdot1+32=2+32=34 \dfrac{g}{mol}$

$\bf \mathcal M_{N_2}=2\cdot A_{N}=2\cdot14=28 \dfrac{g}{mol}$

$\bf \mathcal M_{CuSO_4}=A_{Cu}+A_{S}+4\cdot A_{O}=64+32+4\cdot 16=96+64=160\dfrac{g}{mol}$

$\bf \mathcal M_{Na_2CO_3}=2\cdot A_{Na}+A_{C}+3\cdot A_{O}=2\cdot23+12+3\cdot16=46+12+48=106 \dfrac{g}{mol}$

$\bf \mathcal M_{Al_2(SO_4)_3}=2\cdot A_{Al}+3\cdot A_{S}+12\cdot A_{O}=2\cdot27+3\cdot32+12\cdot 16=54+96+192=342 \dfrac{g}{mol}$

Succes!

-Mihai

$\sf P.S. \: Gliseaza \: spre \: stanga \: daca \: nu \: vezi \:intreg\: raspunsul.$