NaOH + HNO3 -----> NaNO3 + H2O
NaOH + HNO3 -> NaNO3 + H2O
Yes.
The chemical equation is:Sn + 4 HNO3 = SnO2 + 4 NO2 + 2 H2O
Ca(OH)2 + Na2CO3 = CaCO3 + 2 NaOH
IN this reaction the conjugate base is the nitrate ion, NO3-
NaOH + HNO3 -> NaNO3 + H2O
Yes.
HNO3+ NaOH = NaNO3+ H2O is a neutralization reaction
The chemical equation is:Sn + 4 HNO3 = SnO2 + 4 NO2 + 2 H2O
Ca(OH)2 + Na2CO3 = CaCO3 + 2 NaOH
IN this reaction the conjugate base is the nitrate ion, NO3-
Balanced Equation. NaOH + HNO3 >> NaNO3 + H2O Now, Molarity = moles of solute/liters of solution 0.800M HNO3 + mol/2.50L mol of HNO3 = 2 2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH ) = 79.996 grams
Yes. The reaction is fast and exothermic. The equation is NaOH + HNO3 --> H2O + NaNO3
NaNO3 and water
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
Balanced equation first. HNO3 + NaOH >> NaNO3 + H2O Now, Molarity = moles of solute/volume of solution ( find moles HNO3 ) 0.800 M HNO3 = moles/2.50 Liters = 2 moles of HNO3 ( these reactants are one to one, so we can proceed to grams NaOH ) 2 moles NaOH (39.998 grams NaOH/1mole NaOH) = 79.996 grams of NaOH need to neutralize the acid. ( you do significant figures )
NaOH = 40 Mwt so 15/40 moles present. This requires 15/40 moles of HNO3 from the above equation. The HNO3 contains 2 moles in 1000 ml and so 1 mole in 500 ml and therefore 500 x 15/40 = 137.5 mls required