This yields a light sensitive AgCl precipitate
the above answer is not possible since there is no Silver (Ag) in either HNO3 or NaCl
The theoretical yield of HNO3 is the maximum amount of nitric acid that can be produced in a chemical reaction based on stoichiometry and assuming 100% yield. It can be calculated by converting the given amount of reactant to moles, using the balanced equation, and then converting moles to mass of HNO3.
The pH of a 1.45M HNO3 solution is approximately 0.14. This is because nitric acid is a strong acid that dissociates completely in water to yield H+ ions.
HCl + NaOH -> NaCl + H2O is an unbalanced neutralization reaction.
This yields nitric and nitous acid (1 to 1) H2O + 2 NO2 --> HNO2 + HNO3
Extracting NaCl with water three times helps to ensure that as much NaCl as possible is extracted from the solid material. Each extraction helps to increase the efficiency of separating NaCl from impurities or other soluble components. By repeating the process multiple times, the overall yield of NaCl is increased, resulting in a purer final product.
Mixing hydrochloric acid (HCl) and sodium nitrate (NaNO3) will result in the formation of sodium chloride (NaCl) and nitric acid (HNO3). This reaction occurs as a double displacement reaction where the cations and anions switch partners.
When nitric acid (HNO3) reacts with calcium hydroxide (Ca(OH)2), it forms calcium nitrate (Ca(NO3)2) and water (H2O) as products.
To obtain HNO2 from HNO3, you can reduce nitric acid (HNO3) using a reducing agent such as sulfamic acid or a metal like copper. The reduction reaction will convert the nitric acid to nitrous acid (HNO2). Make sure to carry out the reaction in a controlled environment with proper safety measures.
Percentage yield is given as:-%yield = (mass of product obtained experimentally/theoretical yield)x100where the theoretical yield is the maximum mass which could possibly be obtained assuming 100% stoichiometry.So, say as an example if you start with 71g of Cl2 then:-mol = mass/Mr Mr of Cl2 = 71mol = 71/71mol = 1 of Cl2This is the limiting reagent, assuming we have at least 46g of sodium, thus we double this number as the reaction is 1:21 mole Cl2---->2moles NaClagain we use:-mole = mass/Mr Mr of NaCl = 58.52 x Mr = mass2 x 58.5 = massmaximum mass of NaCl possible = 117gIf the reaction is carried out and we achieve 100g of NaCl then:-%yield = (100/117) x 100= 85.5%
The chemical formula for Hydrogen Nitrate is: HNO3
HNO3 is Nitric Acid
1 mg/l = 1ppm1 g/l = 1000 ppmNaCl ----------------> Na+ + Cl-58.5 g NaCl gives 22.989 g Na+(58.5 / 22.989) = 2.545 g of NaCl is required to give 1 g of Na+When 2.545 g of NaCl is dissolved in 1 liter water will yield 1000 ppm of Na+ in NaCl solution.