calculate the no.of o2 atoms in 143 gram of na2co3.10h2o
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Molecular weight of Na2CO3.10H2O
= 286 g
286 g of Na2CO3.10H2O contains
3×6.022×10²³ O2 atoms
hence 1 g of Na2CO3.10H2O contains
3×6.022×10²³/286
0.063×10²³
143 g will contain
0.063×10²³×143
9.009×10²³
= 286 g
286 g of Na2CO3.10H2O contains
3×6.022×10²³ O2 atoms
hence 1 g of Na2CO3.10H2O contains
3×6.022×10²³/286
0.063×10²³
143 g will contain
0.063×10²³×143
9.009×10²³
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