Calculating the percentage of silica

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Discussion Overview

The discussion revolves around calculating the percentage of silica in a partially dried clay mineral sample, given the original composition of the sample, including water and silica content. The context is primarily homework-related, focusing on the application of mathematical reasoning to solve the problem.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant states the original sample contains 12% water, 45% silica, and 43% other components, suggesting a total mass of 100 g for ease of calculation.
  • Another participant reiterates the initial breakdown of the sample and prompts for further calculations regarding the water content after drying.
  • A third participant expresses difficulty in progressing with the calculations despite understanding the initial setup.
  • There is a suggestion to define the remaining water content after drying as 'x' and to maintain the silica and other components constant for the calculation.

Areas of Agreement / Disagreement

Participants generally agree on the initial composition of the sample and the approach to calculating the percentage of silica, but there is no consensus on the specific calculations or the next steps to take.

Contextual Notes

Limitations include the lack of clarity on how to handle the change in water content and its effect on the overall percentage of silica in the partially dried sample. The discussion does not resolve the mathematical steps needed to arrive at a final answer.

xiphoid
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Homework Statement


A partially dried clay mineral contains 8% of water, the original sample contained 12% water and 45% silica. Then, the percentage of silica in the partially dried sample is nearly?


Homework Equations


Don't know if this is would help in the calculation,
in the original sample, 12% water is present, hence non water components are consisting of 88%


The Attempt at a Solution


Attempted one but was out of range of the concept to be used.
 
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hi xiphoid! :smile:

hint: suppose the original sample was 100 g (12 g water, 45 g silica, 43 g other stuff) :wink:
 
I had that stuff in my mind but couldn't manage to calculate further
tiny-tim said:
hi xiphoid! :smile:

hint: suppose the original sample was 100 g (12 g water, 45 g silica, 43 g other stuff) :wink:
 
xiphoid said:
I had that stuff in my mind but couldn't manage to calculate further

ok, after the sample has dried, the water is less but everything else is the same

so call it x g water, 45 g silica, 43 g other stuff, …

from the question, what equation can you find for x ? :wink:
 

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