This depends on the mass of calcium carbonate.
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
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.
To find the molarity, first calculate the number of moles of HNO3: 0.31g / 63g/mol (molar mass of HNO3) = 0.0049 mol Then, convert 300ml to liters: 300ml/1000 = 0.3 L Finally, molarity = moles/volume = 0.0049 mol / 0.3 L = 0.0163 M
One equivalent proton per mole, so molarity is equal to normality.
HNO3 is a strong acid, which means it dissociates completely. This means you don't have to set up an equilibrium scenario; you can just go with the given molarity as also being the concentration of hydrogen ions [H+]. So, pH = -log(0.00884), which is about 2.05.
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.
When CaCO3 is added to HNO3, a chemical reaction occurs where CaCO3 reacts with HNO3 to produce Ca(NO3)2, CO2, and H2O. This reaction is a double displacement reaction where the calcium ions in CaCO3 switch places with the nitrate ions in HNO3.
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.
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
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.
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).
The molarity of nitric acid, HNO3, can vary depending on the concentration of the solution. For example, a 1 M solution of nitric acid would contain 1 mole of HNO3 per liter of solution. It is important to know the concentration or volume of the solution to determine the molarity of nitric acid.
For the reaction between HNO3 (acid) and KOH (base), it is a 1:1 molar ratio reaction. This means that 1 mole of HNO3 will react with 1 mole of KOH. So, 1 mole of KOH is required to neutralize 1 mole of HNO3 in this reaction.
To find the molarity, first calculate the number of moles of HNO3: 0.31g / 63g/mol (molar mass of HNO3) = 0.0049 mol Then, convert 300ml to liters: 300ml/1000 = 0.3 L Finally, molarity = moles/volume = 0.0049 mol / 0.3 L = 0.0163 M
No. HNO3 already has hydrogen and nitrogen in their highest possible oxidation states.
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.
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.