0
What else can I help you with?
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.
How many moles of potassium hydroxide are needed to neutralize 3 moles of nitric acid?
The balanced chemical equation for the reaction between potassium hydroxide (KOH) and nitric acid (HNO3) is 1:1 ratio. Therefore, 3 moles of nitric acid will require 3 moles of potassium hydroxide to neutralize it.
What mass of limestone in kg would be required to completely neutralize a 15.4 billion-liter lake that is 1.9105 M in H2SO4 and 8.7106 M in HNO3?
To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.
How many moles of sodium hydroxide are required to neutralize 20 mol of nitric acid?
The balanced chemical equation for the neutralization between sodium hydroxide (NaOH) and nitric acid (HNO3) is 1 mol of NaOH reacts with 1 mol of HNO3. Therefore, 20 moles of nitric acid would require 20 moles of sodium hydroxide to neutralize it.
How many mL of 0.4040 M HNO3 would be needed to neutralize 55.00 mL of 0.1234 M potassium hydroxide?
To determine the volume of HNO3 required to neutralize KOH, you can use the concept of stoichiometry. The mole ratio between HNO3 and KOH is 1:1. Calculate the moles of KOH from its molarity and volume, then use the mole ratio to find the moles of HNO3 needed. Finally, convert the moles of HNO3 to volume using its molarity, which will give you the milliliters needed.
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.
How many moles of potassium hydroxide are needed to neutralize 3 moles of nitric acid?
The balanced chemical equation for the reaction between potassium hydroxide (KOH) and nitric acid (HNO3) is 1:1 ratio. Therefore, 3 moles of nitric acid will require 3 moles of potassium hydroxide to neutralize it.
What mass of limestone in kg would be required to completely neutralize a 15.4 billion-liter lake that is 1.9105 M in H2SO4 and 8.7106 M in HNO3?
To calculate the amount of limestone needed, first determine the moles of H2SO4 and HNO3 in the lake. Then, use the stoichiometry of the neutralization reaction between limestone (CaCO3) and the acids to calculate the moles of limestone required to neutralize the acids. Finally, convert the moles of limestone to kilograms using the molar mass of CaCO3.
How many moles of sodium hydroxide are required to neutralize 20 mol of nitric acid?
The balanced chemical equation for the neutralization between sodium hydroxide (NaOH) and nitric acid (HNO3) is 1 mol of NaOH reacts with 1 mol of HNO3. Therefore, 20 moles of nitric acid would require 20 moles of sodium hydroxide to neutralize it.
How many moles of HNO3 are required to react with 6 moles of Cu?
The balanced chemical equation for the reaction between copper (Cu) and nitric acid (HNO3) is: 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O According to the equation, 8 moles of HNO3 are required to react with 3 moles of Cu. Therefore, to react with 6 moles of Cu, you would need 16 moles of HNO3.
How many mL of 0.4040 M HNO3 would be needed to neutralize 55.00 mL of 0.1234 M potassium hydroxide?
To determine the volume of HNO3 required to neutralize KOH, you can use the concept of stoichiometry. The mole ratio between HNO3 and KOH is 1:1. Calculate the moles of KOH from its molarity and volume, then use the mole ratio to find the moles of HNO3 needed. Finally, convert the moles of HNO3 to volume using its molarity, which will give you the milliliters needed.
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 )
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 volume in drops of 16M HNO3 is required to react with 0.0214grams of Cu metal?
To find the volume of 16M HNO3 required to react with 0.0214g of Cu metal, you need to calculate the moles of Cu. Then, using the balanced equation for the reaction between Cu and HNO3 (Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O), you can determine the moles of HNO3 needed. Finally, using the molarity of the HNO3 solution, you can calculate the volume in drops.
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
A 3.0-milliliter sample of hno3 solution is exactly neutralized by 6.0 milliliters of 0.50 m koh what is the molarity of the hno3 sample?
The moles of KOH can be calculated as (0.50 mol/L) x (6.0 mL). Since KOH is in a 1:1 ratio with HNO3 in the neutralization reaction, the moles of HNO3 are the same as KOH. So, the molarity of the HNO3 sample would be (moles of HNO3) / (3.0 mL).
How many moles of base are needed to neutralize 2 mol of acid for HNO3 plus KOH?
Since both the acid and the base have equivalent weights equal to their formula weights, 2 moles of KOH are needed to neutralize 2 moles of nitric acid.
