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What is the acid base reaction for HNO3 plus NaOH?
The acid-base reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) produces sodium nitrate (NaNO3) and water (H2O). The balanced chemical equation for this reaction is: HNO3 + NaOH → NaNO3 + H2O.
What is the conjugate base for the equation hno3 plus naoh h2o plus nano3?
The conjugate base for the equation HNO3 + NaOH -> H2O + NaNO3 is the nitrate ion (NO3-), which is formed when the strong acid HNO3 donates a proton to water (H2O) to form the weak conjugate base NO3-.
What is the molarity of HNO3 if 20.0 ml of the solution is needed to exactly neutralize 10.0 ml of a 1.67 M NaOH solution?
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
Is heat generated with NaOH plus HNO3 reaction?
Yes, heat is typically generated when NaOH reacts with HNO3 due to the exothermic nature of the reaction. This reaction produces sodium nitrate (NaNO3) and water (H2O) as products along with heat.
How do i balance this equation HNO3 plus Sn ------ SnO2 plus HNO2 plus H2O?
The chemical equation is:Sn + 4 HNO3 = SnO2 + 4 NO2 + 2 H2O
What is the acid base reaction for HNO3 plus NaOH?
The acid-base reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) produces sodium nitrate (NaNO3) and water (H2O). The balanced chemical equation for this reaction is: HNO3 + NaOH → NaNO3 + H2O.
HNO3 + NaOH NaNO3 + H2O?
HNO3+ NaOH = NaNO3+ H2O is a neutralization reaction
What is the conjugate base for the equation hno3 plus naoh h2o plus nano3?
The conjugate base for the equation HNO3 + NaOH -> H2O + NaNO3 is the nitrate ion (NO3-), which is formed when the strong acid HNO3 donates a proton to water (H2O) to form the weak conjugate base NO3-.
What is the molarity of HNO3 if 20.0 ml of the solution is needed to exactly neutralize 10.0 ml of a 1.67 M NaOH solution?
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
Is heat generated with NaOH plus HNO3 reaction?
Yes, heat is typically generated when NaOH reacts with HNO3 due to the exothermic nature of the reaction. This reaction produces sodium nitrate (NaNO3) and water (H2O) as products along with heat.
How do i balance this equation HNO3 plus Sn ------ SnO2 plus HNO2 plus H2O?
The chemical equation is:Sn + 4 HNO3 = SnO2 + 4 NO2 + 2 H2O
What is the molarity of a nitric acid solution HNO3 if 20.0 mL of the solution is needed to exactly neutralize 10.0 mL of a 1.67 M NaOH solution?
The balanced chemical equation for the neutralization reaction is: HNO3 + NaOH → NaNO3 + H2O From the equation, we know that the mole ratio of HNO3 to NaOH is 1:1. This means that the moles of HNO3 will be equal to the moles of NaOH. First, calculate the moles of NaOH used: Moles NaOH = (10.0 mL) x (0.001 L/mL) x (1.67 mol/L) = 0.0167 mol Since the moles of NaOH are equal to the moles of HNO3, we have 0.0167 mol of HNO3 in 20.0 mL of solution: Molarity of HNO3 = moles HNO3 / volume of solution (L) = 0.0167 mol / 0.020 L = 0.835 M
What is the molarity of an naoh solution if 4.37ml is titrated by 11.1 ml of 0.0904 m hno3?
The balanced equation for the reaction is 1 mole of NaOH to 1 mole of HNO3. Using the titration data, you can calculate the moles of HNO3 used. From there, you can determine the moles of NaOH present in the 4.37 ml solution. Finally, dividing the moles of NaOH by the volume of the NaOH solution in liters will give you the molarity.
How many moles of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?
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
What is the equation of a reaction of nitric acid and sodium hydroxide?
HNO3 (aq) + NaOH (aq) --> H2O (l) + NaNO3 (aq)
What is the balanced equation for the reaction of nitric acid and sodium hydroxide?
Nitric acid: HNO3 (acid) Sodium hydroxide: NaOH (base) This is therefore an acid-base reaction. Acid + Base --> Salt + Water Therefore: HNO3 + NaOH --> NaNO3 + H20 Or: Nitric acid + Sodium hydroxide --> Sodium Nitrate + Water
A nitric acid soulution is neutralized using sodium hydroxide how many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M Nitriric acid soulution?
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 )
