NaBr is a salt because it has Na in it
CuNo32 may be a nonelectrolyte
NH3 is a base
KOH is a base
CaCl2 may be a nonelectrolyte
hope that somewhat helps
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
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
Yes all sodium (Na) compounds are water soluble.
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.
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
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.
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%.
Sodium nitrate (NaNO3) is a neutral salt. When dissolved in water, it will not significantly affect the pH of the solution.
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.
The molarity is 5,55.
A reaction doesn't exist; the solution contain ions of calcium, iodine, sodium and nitrate.
No.Since,sodium nitrite(NaNO2)is a salt of strong base and weak acid,in water solution it will leave more amount of [OH-] ions.Hence, solution will be basic.Therefore,NaNO2 is base not an acid.
This percentage is 12 %.
To calculate the amount of NaNO3 needed to form a saturated solution at 45°C, the solubility of NaNO3 at this temperature must be known. Once this value is determined, the grams of NaNO3 required can be calculated using the solubility value and the saturation point equation.
The pH of a 1.0 M aqueous solution of NaNO3 will be around 7, which is considered neutral. Sodium nitrate is a strong electrolyte that dissociates completely into Na+ and NO3- ions, neither of which will affect the pH of the solution significantly.
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