0.499g 0f CuSO₄.xH₂O when heated to constant weight gave a residue of 0.346g. the value of x is? (Cu = 63.5, S = 32.0, O = 16, H = 1)

olyezema

4 years ago

CuSO4/CuSO4.H2O= mass of CuSO4/mass of CuSO4.H2O
159.5/159.5+18x= 0.346/0.499
criss multiple
159.5×0.449=159.5×0.346+18x×0.346
79.59=55.2+6.2x
79.59-55.2=6.2x
22.39/6.2=x
x=3.9~4......D

Somysoph

5 years ago

Mass of anhydrous salt/molar mass of anhydrous salt=mass of hydrated salt /molar mass of hydrated salt.
0.346/159.5=0.499/159.5+18x
Cross multiply and your and will be 3.92 approximately 4.0. To calculate mass of anhydrous salt do not add mass of xH2O only mass of cuso4

anihfriday

6 years ago

Explain pls

Gracefularc

6 years ago

explanation please

tamunoimamadeinma

2 months ago

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Sheriffoo1

2 months ago

i don't understand??

Simply..Lee

3 months ago

i dont get

bostyt

3 years ago

4