Molarity is mols of solute over liters of solution.
85.0g NaNO3 (1mol/85g ) = 1mol
750ml = 0.75 Liters
so,
1/0.75 = 1.3 Molar
First, convert the mass of NaNO3 to moles using the molar mass of NaNO3. Then, calculate the molarity by dividing the moles of NaNO3 by the volume of the solution in liters. Molarity (M) = moles of solute / volume of solution in liters.
4.25 g NaNO3 x 1 mole NaNO3/85 g = 0.05 moles0.05 moles/0.1 L = 0.5 moles/L = 0.5 M
The total mass of the solution is 4.2g NaNO3 + 60g water = 64.2g. The percent concentration by mass of the NaNO3 in the solution is (4.2g / 64.2g) * 100% = 6.54%.
To find the moles of solute, multiply the molarity by the volume in liters. So, (0.75 , \text{mol/L} \times 1.25 , \text{L} = 0.9375 , \text{moles of NaNO}_3).
Yes, sodium nitrate (NaNO3) is highly water soluble. It dissolves easily in water to form a clear, colorless solution.
NaNO3, or sodium nitrate, is a neutral salt when dissolved in water because it is a strong electrolyte that dissociates completely into sodium ions and nitrate ions. The presence of both cations (sodium ions) and anions (nitrate ions) in solution makes the overall solution neutral.
The molarity is 5,55.
To find the moles of solute, multiply the volume of the solution in liters (1.25 L) by the molarity of the solution (0.75 mol/L). Therefore, the moles of NaNO3 in 1.25 L of 0.75M solution is 0.9375 moles.
4.25 g NaNO3 x 1 mole NaNO3/85 g = 0.05 moles0.05 moles/0.1 L = 0.5 moles/L = 0.5 M
The mass percent of a saturated solution of NaNO3 at 20°C can be calculated by dividing the mass of NaNO3 in the saturated solution (82 g) by the total mass of the solution (82 g NaNO3 + 100 g water) and then multiplying by 100%. This results in a mass percent of 82/182 * 100% ≈ 45.1%.
NaNO3 contains ionic bonds between Na+ and NO3-, while C2H3OH contains both covalent and ionic bonds due to the presence of both carbon-carbon and carbon-oxygen bonds; CH3Cl contains a covalent bond between carbon and chlorine; NH2OH has covalent bonds between nitrogen and hydrogen, as well as nitrogen and oxygen; H2O2 contains covalent bonds between hydrogen and oxygen; CH3C likely refers to CH3COOH (acetic acid), which contains covalent bonds between carbon, hydrogen, and oxygen.
Sodium nitrate (NaNO3) is a neutral salt. When dissolved in water, it will not significantly affect the pH of the solution.
Molarity (M) = moles of solute (mol) / liter of solution (L)M = mol / LYou have 250 mL of Solution, which is250 mL x ( 1 L / 1000 mL ) = ( 250 / 1000 ) L = .25 LSolute is just what's dissolvedSolvent is just what it's being dissolved inSolution is the solute and the solvent.M = mol / LM = 0.65 mol / 0.25 Liters = 2.6 mol/LThe two numbers that you are given, 0.65 moles and 250 mL both have two significant figures, and the answer is two significant figures (2.6 mol/L)Therefore the answer is 2.6 mol/L.
To find the moles of solute, multiply the molarity by the volume in liters. So, (0.75 , \text{mol/L} \times 1.25 , \text{L} = 0.9375 , \text{moles of NaNO}_3).
To neutralize the nitric acid, you need a 1:1 mole ratio of sodium hydroxide to nitric acid. First, calculate the moles of nitric acid in the solution using the formula Molarity = moles/volume. Then, use the mole ratio to find the moles of sodium hydroxide needed. Finally, convert this to grams using the molar mass of sodium hydroxide.
To find the volume of AgNO3 solution needed, we first calculate the moles of NaCl using the molar ratio between AgNO3 and NaCl in the balanced chemical equation. Then, we use the molarity of the AgNO3 solution to determine the volume: moles of NaCl x (1 mol AgNO3 / 1 mol NaCl) x (1 L / 0.117 mol) x 1000 mL = volume in mL.
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
The total mass of the solution is 4.2g NaNO3 + 60g water = 64.2g. The percent concentration by mass of the NaNO3 in the solution is (4.2g / 64.2g) * 100% = 6.54%.