A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heating cycle, the mass of the sample and the dish is 13.266 g. The student stops the heat/cool/weigh cycle after the second time. If the original mass of the evaporating dish and the hydrate was 25.637 g, and the mass of the evaporating dish alone was 1.135 g, what is the mass of water removed from the sample by heating? Provide your response to three digits after the decimal.

Question

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Respuesta :

Mergus Mergus · Answer 1 · 2019-10-24T05:32:01+02:00

Answer:

12.371 g

Explanation:

Given :

[tex]m_{evaporating\ dish}=1.135\ g[/tex]

[tex]m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g[/tex]

[tex]m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g[/tex]

[tex]m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g[/tex]

Mass of salt hydrate:

[tex]m_{evaporating\ dish}=1.135\ g[/tex]

[tex]m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g[/tex]

[tex]m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g[/tex]

Mass of salt anhydrous:

[tex]m_{evaporating\ dish}=1.135\ g[/tex]

[tex]m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g[/tex]

[tex]m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g[/tex]

Mass of water:

[tex]m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g[/tex]

[tex]m_{water}=12.371\ g[/tex]