d r e a m y
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Hi, I’m not sure if I did these right. Looking at the last question, what density was I supposed to use? liquid water or ice?
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louisayapp
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Hello - hope you're well
In answer to your problem - I would take this approach!
use the density of the ice to find the mass of the H2O present, then convert this mass into a volume for liquid water;

3*0.917 = 2.751g of H2O (volume x density)
2.751/1 = 2.752cm^3 (mass/density)

hope that helps
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menzie91
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This is right.
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d r e a m y
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(Original post by louisayapp)
Hello - hope you're well
In answer to your problem - I would take this approach!
use the density of the ice to find the mass of the H2O present, then convert this mass into a volume for liquid water;

3*0.917 = 2.751g of H2O (volume x density)
2.751/1 = 2.752cm^3 (mass/density)

hope that helps
thank you!!
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louisayapp
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(Original post by d r e a m y)
thank you!!
That's okay
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