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ca=40
2br=160
40:160=1:4
b)
fe2=112
o3=48
112:48=2,333:1
2a)so3=32+3*16=80u
s=80:32=40%
03=100%-40%=60%
b)al2s3=2*27+3*32=150
al2=150:54=36%
s3=100%-36%=64%
a)CaBr2
mCa : mBr = 40u : 2*80u
mCa : mBr = 40 : 160
mCa : mBr = 1 : 4
b)Fe2O3
mFe : mO = 2*56u : 3*16u
mCa : mBr = 112 : 48
mCa : mBr = 7 : 3
2.
a)tlenku siarki(VI)
mSO3=32u+3*16u=80u
80g SO3--------100%
32g siarki------x%
x=32*100/80
x=40% siarki
80g SO3--------100%
48g tlenu------x%
x=48*100/80
x=60% tlenu
b)siarczku glin
mAl2S3=2*27u+3*32u=150u
150g Al2S3------100%
54g glinu--------x%
x=54*100/150
x=36% glinu
150g Al2S3------100%
96g siarki--------x%
x=96*100/150
x=64% siarki