Question

Policz masy cząsteczkowe :

mH2S =

mAgNO3=

mk2SO4=

mAL2(CO3)3=

mH3PO4=

mFe(NO3)=

ma=

mliOH=

mNH3=

mMg(OH)2=

Answer 1

m H₂S = 1g *2 + 32g = 34g 

m AgNO₃ = 108g + 14g + 3 * 16g = 170g

m K₂SO₄ = 2 * 39g + 32g + 4 * 16g = 174g 

m Al₂(CO₃)₃ = 2 * 27g + 3 * 12g + 3 * 3 * 16g = 54g + 36g + 144g = 234g 

m H₃PO₄ = 3 * 1g + 31g + 4 * 16g = 98g 

m Fe(NO₃)₂ = 56g + 2 * 14g + 2 * 3 * 16g = 180g 

m Fe(NO₃)₃ = 56g + 3 * 14g + 3 * 3 * 16g = 242g 

ma? Czyli? 

m LiOH = 7g + 16g + 1g = 24g 

m NH₃ = 14g + 3 * 1g = 17g 

m Mg(OH)₂ = 24g + 2 * 16g + 2 * 1g = 58g 

Answer 2

m H₂S = 2*1u + 32u = 34u

 

m AgNO₃ = 108u + 14u + 3*16u = 170u

 

m K₂SO₄ = 2*39u + 32u + 4*16u = 174u

 

m Al₂(CO₃)₃= 2*27u + 3*(12u + 3*16u) = 234u

 

m H₃PO₄= 3*1u + 31u + 4*16u = 98u

 

Dla azotanu (V) żelaza (III) : m Fe(NO₃)₃= 56u + 3*(14u + 3*16u) = 242u

lub :

Dla azotanu (V) żelaza (II) : m Fe(NO₃)₂ = 56u + 2*(14u + 3*16u) = 180

 

ma = tu niestety nic nie wymyślę...

 

m LiOH = 7u + 16u + 1u = 24u

 

m NH₃ = 14u + 3*1u = 17u

 

m Mg(OH)₂ = 24u + 2*(16u + 1u) = 58u