Question

1. Oblicz masy cząsteczkowe podanych związków chemicznych:

mCuO = .......................

mFe₂O₃ = .....................

mNa₂CO₃ = ....................

mH₂SO₄ = ......................

mCa(OH)₂ = ...................

mFe₂(SO₃)₃ = ..................

Answer 1

M CuO = 16u + 63,5u = 79,5u

m Fe₂O₃ = 3*16u + 2 * 56u = 48u + 112u = 160u

m Na₂CO₃ = 2*23u + 12u + 3*16u = 106u

m H₂SO₄ = 2*1u + 32u + 4*16u = 98u

m Ca(OH)₂ = 40u + 2*16u + 2*1u = 74u

m Fe₂(SO₃)₃ =2*56u + 3*32u + 9*16u = 112u + 96u + 144u = 352u

Answer 2

MCu + mO= 64 u + 16 u = 80 u
2mFe + 3mO = 2*56 u + 3*16 u = 160 u
2mNa + mC + 3mO = 2* 23 u + 12 u + 3*16 u = 106 u
2mH + mS+ 4mO = 2*1 u + 32u +4*16 u = 98 u
mCa + 2mO + 2mH= 40 u + 2*16 u + 2*1 u = 74 u
2mFe +3mS +9mO = 2*56 u + 3*32 u + 9*16 u = 352 u

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