In a 0.4g sample of Mg3(PO4)2, how many phosphorus atoms would you find
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262g of mg3(po4)2 contain 2×6.023×10^23 phosphorus atoms.
Now,
0.4g of mg3(po4)2 contain
=(0.4×2×6.023×10^23)/262
=0.0184×10^23
=184×10^19atoms of phosphorus.
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Now,
0.4g of mg3(po4)2 contain
=(0.4×2×6.023×10^23)/262
=0.0184×10^23
=184×10^19atoms of phosphorus.
Hope it helps!!
Pls mark it as the brainliest.
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