Calculating Coulombs of Charge in Water

AI Thread Summary
To calculate the number of coulombs of positive charge in 250 cm³ of neutral water, the initial step involves converting the volume to moles using water's density, resulting in approximately 13.87 moles of H₂O. This equates to about 8.35 x 10²⁴ water molecules, which contains a total of 8.35 x 10²⁵ protons (2 from each hydrogen atom and 8 from the oxygen atom). Multiplying the total number of protons by the charge of a proton (1.6 x 10⁻¹⁹ Coulombs) yields approximately 1,336,736 Coulombs. The calculations for electrons match this value, indicating a balance between positive and negative charges in neutral water. The overall conclusion confirms the accuracy of the calculations despite initial concerns about the large number.
Valhalla
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The problem states: Calculate the number of coulombs of positive charge in 250 cm3 of (neutral) water. (Hint: A hydrogen atom contains one proton and an oxygen atom contains eight protons.)

Ok, first I converted to moles. Using the density of water.

250mL * \frac{1g}{1mL} * \frac{1 mole H_2O}{18.02g}=13.8735moles H_2O

13.8735molesH_2O*6.022*10^{23}=8.3546*10^{24} atoms H_2O

8.3546*10^{24}*(2*(1 proton H)+8 proton O)=8.3546*10^{25} protons

8.3546*10^{25} protons * 1.6*10^{-19} Coulombs=1336736Coulombs

Does this seem right? That seems really large. However, the electrons calculate up to exactly the same amount and cancel it out.
 
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oh ok, I didn't check the second page. Thanks!
 

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