Put the elements in order from the highest electronegativity at the top of your list, to the lowest at the bottom. Germanium (Ge
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We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
![[ OH^{-}]=3.31* 10^{-7} M](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D3.31%2A%2010%5E%7B-7%7D%20M%20%20)
Then we can transform the equation into:
Then we are able to plug in the pOH and directly get [OH-]: