oblicz mase NH3
K20
NaoH
NH3
N=14u
H=1u
m NH3=14u + 3*1u = 14u + 3u= 17u
K2O
K=39u
O=16u
m K2O= 2*39u + 16u= 78u + 16u= 94u
NaOh
Na=23u
m NaOH= 23u + 16u + 1u= 40u
mNH3=14u+3*1u=17u
mK2O=2*39u+16u=94u
mNaOH=23u+16u+1u=40u
masa NH₃
mN= 14g
mH= 1g
mNH₃= 14g+ 3*1g= 14g+ 3g= 17g
masa K₂O
mK= 39g
mO=16g
mK₂O= 2*39g+ 16g= 78g+ 16g= 94g
masa NaOH
mNa=23g
mH=1g
m NaOH= 23g+ 16g+ 1g= 40g1g=1uPozdrawiam, kate66 ;)
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NH3
N=14u
H=1u
m NH3=14u + 3*1u = 14u + 3u= 17u
K2O
K=39u
O=16u
m K2O= 2*39u + 16u= 78u + 16u= 94u
NaOh
Na=23u
O=16u
H=1u
m NaOH= 23u + 16u + 1u= 40u
mNH3=14u+3*1u=17u
mK2O=2*39u+16u=94u
mNaOH=23u+16u+1u=40u
masa NH₃
mN= 14g
mH= 1g
mNH₃= 14g+ 3*1g= 14g+ 3g= 17g
masa K₂O
mK= 39g
mO=16g
mK₂O= 2*39g+ 16g= 78g+ 16g= 94g
masa NaOH
mNa=23g
mO=16g
mH=1g
m NaOH= 23g+ 16g+ 1g= 40g
1g=1u
Pozdrawiam, kate66 ;)