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When diluting a mixture of two reactants with lets say a 100ml of H2O, how do you add this to the uncertainties of the concentrations of the reactants? Should the relative uncertainty of the added H2O be added to the relative uncertainty of the volume in both reactants?

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When adding values together, you should add together their absolute uncertainties. For example, if you started with $500\pm5\mathrm{mL}$ and add $100\pm1\mathrm{mL}$, your new volume value is $600\pm6\mathrm{mL}$.

You would add the relative uncertainties if you are multiplying values. For example, if you started with $1\pm.06\text{ moles}$ and want to find the molarity in a solution with a volume of $500\pm5\mathrm{mL}$, your molarity will be $2\pm.14\text{M}$. The uncertainty was obtained by finding the sum of the relative uncertainties and multiplying it by the concentration to obtain an absolute uncertainty, as shown below. $$(\frac{.06}{1}+\frac{5}{500})(2\text{M})=\frac{7}{100}(2\text{M})=.14\text{M}$$

For more details on uncertainty rules, this page from the University of Victoria gives some of the basics.

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When adding or subtracting uncertainties they are squared and added then the square root taken. Thus if $z=x+y$ where x and y are two measurements with uncertainties $\sigma_x, \sigma_y$ the combined uncertainty is $$\sigma_z = \sqrt{\sigma _x^2 + \sigma_y^2}$$ and the answer $z= (x+y) \pm \sigma_z$

For more details besides adding values see this answer Uncertainty Equations Confusion

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  • $\begingroup$ @L. Wagemaker I would recommend this answer. My method is a simplification, which is easier to compute, but will typically overestimate the uncertainty. $\endgroup$ – Tyberius Apr 1 '17 at 23:24

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