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If 4.25 g of NaNO3 (molar mass 85 gmole) are dissolved in 100 mL of H2O. What is the molarity of the resulting solution?
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
What is the percent concentration by mass of a solution 4.2g of NaNO3 in 60 g water?
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
How many moles of solute are present in 1.25L of a 0.75M NaNO3?
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).
Is nano3 water soluble?
Yes all sodium (Na) compounds are water soluble.
NaNo3 is it acidic basic or neutral when dissolved in water?
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.
What is the molarity of a solution that has 2.5 moles of NaNO3 in 0.45 L?
The molarity is 5,55.
How many moles of solute are present in 1.25 L of a 0.75M NaNO3 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.
If 4.25 g of NaNO3 (molar mass 85 gmole) are dissolved in 100 mL of H2O. What is the molarity of the resulting solution?
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
What is the percent concentration by mass of a solution 4.2g of NaNO3 in 60 g water?
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
The solubility of NaNO3 at 20 C is 82 g per 100 grams of water What is the mass percent of a saturated solution of NaNO3?
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%.
Is NaNO3 acidic neutral or basis?
Sodium nitrate (NaNO3) is a neutral salt. When dissolved in water, it will not significantly affect the pH of the solution.
What is the molarity of a solution that contains 0.65 moles of NaNO3 in 250 mL of 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.
How many moles of solute are present in 1.25L of a 0.75M NaNO3?
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).
A nitric acid solution is neutralized using sodium hydroxide How many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M nitric acid solution?
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.
How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can
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 is one of the possible products of a solution if CaI2 is mixed with a solution of NaNO3?
A reaction doesn't exist; the solution contain ions of calcium, iodine, sodium and nitrate.
