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Because it is based on conservation of moles. M1V1 = moles before dilution = M2V2 = moles after dilution M 1 V 1 = moles before dilution = M 2 V 2 = moles after dilution. As Satwik has stated in the comments, the equation relies on the conservation of moles principle. mol1 V1 ×V1 = mol2 V2 ×V2 m o l 1 V 1 × V 1 = m o l 2 V 2 × V 2.
$$\ce{N1V1 = N2V2}$$ where, N is the Normality of the Acid / Base & V is the volume in litres. We ask you to give a scientific justification or derivation to the above equation i.e.
When calculating the concentration after dilutions, keep in mind the amount of moles is still the same, so just use the formula $\ce{M1V1 = M2V2}$. Also, sodium acetate is a soluble salt (see solubility rules), so in the net ionic equation, don't write out the sodium as it is a spectator ion.
The first step of the answer is converting the given weight of NaX2COX3 ⋅ 10HX2O into amount of substance. For that we have a formula as amount of substance = mass of substance molecular mass of the substance amount of substance = 1 g NaX2COX3 ⋅ 10HX2O 286 g NaX2COX3 ⋅ 10HX2O = 0.003447 mol. We used this first step to determine the amount ...
If you need a more diluted solution, you may use the formula: = M V = M V. where, M1 M 1 and M2 M 2 are the initial and final concentrations, V1 V 1 is the volume of stock you would require for the dilution, and V2 V 2 is the volume of final desired concentration. Let us assume you require 50 mL 50 m L of 10 mM 10 m M solution, and you have a ...
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$\begingroup$ Nitric acid is like the question said just the solvent, so you are on the right track with your formula, as it states before any reaction. $\endgroup$ – Martin - γγΌγγ³ ♦ Commented Aug 30, 2015 at 10:50
1 Answer. Sorted by: The first name is correct (1,1,1-tribromo...). Counting the substituents from either end, you would get 1 for the first substituent going from either direction. As a tie-breaker, look at the next substituent, which would give you 1,1 going from the bromine end, and 1,1 going from the chlorine end.
Actually, you can. HX2SOX4 liberates 2HX + ions which is neutralized by the OHX − ions. The Ka for water is 10 − 14, this means that the concentration of those ions are very low compared to the KX + and the SOX4X2 − ions. You just need to use the stoichiometry of the equation properly to apply it to MaVa = MbVb.