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How many moles of sodium nitrate are in 650 mL of a 0.28 M in NaNo3 solution?
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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 molarity of a solution that contains 4.53 moles of lithium nitrate in 2.85 liters of solution?
This molarity is 1,59.
How many moles of nitrogen N are in 60.0 g of nitrous oxide N2O?
For this you need the atomic (molecular) mass of NaNO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaNO3=85.0 grams60.1 grams NaNO3 / (85.0 grams) = .707 moles NaNO3
What is the concentration of 10. moles of copper (II) nitrate in 5.0 liters of solution?
The concentration is 2 mol/L.
A volume of 10 ml of a .00600 M solution are reacted with 0.500 m solution of AgNO3 what is the maximum mass of AgCL that precipitates?
This was a fun one.Step 1: You're not given the other reactant solution or the other product solution, so let's make one up that works. If you use sodium chloride on the reactant side and sodium nitrate on the product side, you get this correctly balanced double replacement reaction equation:AgNO3(aq) + NaCl(aq) --> NaNO3(aq) + AgCl(s)This works well because you don't need any molar ratio coefficients to muddy things up any further.Step 2: You're given the volume of the unnamed solution (that we've decided will be NaCl) so let's figure out how many moles are dissolved in that volume. Molarity is moles per liter, and you have 10mL, which is 0.01L, so move the decimal point twice to the left to get the moles per 10mL. You should get 0.006 moles of NaCl.Step 3: You're given the molality (m) of the silver nitrate solution, but not its volume. In a weak solution, molality (m) is approximately the same as molarity (M), but it doesn't matter since you aren't given its volume. You should assume then, in this case, that you have an excess of silver nitrate solution, which means that the sodium chloride solution is your limiting reactant. This means that you should now run the stoichiometry with the mass of NaCl.Step 4: The equation above is balanced as is, with no coefficients necessary, so the moles of the NaCl equal the moles of the AgCl, which we established in step 2 was 0.006 moles.Step 5: The precipitate is silver chloride, AgCl, with a molar mass of 143g/mol. Multiply this by the number of moles, 0.006, to get 0.858g AgCl, which is the answer. This is the maximum amount of silver chloride which can be yielded by this reaction under the conditions given.
What is the mass of 6.0 moles of sodium nitrate (NaNO3)?
1 mole of Na2SO4 is 142.05 or 142.1 when using 4 significant figures
How do you calculate the volume in milliliters of a 0.61M sodium nitrate solution that contains 400mmol of sodium nitrate?
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
If 156 g of sodium nitrate react and 112 grams of sodium nitrate are recovered what is the percentage yield?
Moles NaNO3 = 156 g / 84.9965 g/mol = 1.84 we would get 1.84 moles of NaNO2 => 1.84 mol x 68.9965 g/mol = 127 g % = 112 x 100 / 127 = 88.2
How many MOLES of sodium nitrate are present in 2.85 grams of this compound?
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
How many moles of silver nitrate are needed to completely react with 0.327 mole of sodium cloride?
The balanced equation for the reaction is AgNO3 (aq) + NaCl (aq) -> AgCl (s) + NaNO3. The coefficient of each reactant is the implied 1 when no explicit coefficient is shown in the equation. Therefore the same number of moles of silver nitrate as of sodium chloride are required for the reaction.
What is the mass of 1 mol of NaNO3?
Using atomic weights for Na = 23 and N=14 and O = 16, one arrives at a mass for 1 mole of NaNO3 of23 + 14 + (3x16) = 85 grams/mole
What is the molarity of a solution that has 2.5 moles of NaNO3 in 0.45 L?
The molarity is 5,55.
Is the moles of silver ions in water equal to silver nitrate solution moles?
no
How do you Calculate how much sodium nitrate is needed to give 126g of nitric acid by reaction?
The reaction would be NaNO3 (s) + H2SO4 (l) -> HNO3 (l) + NaHSO4 (s)Moles ratio 1 : 1 : 1 : 1Moles of nitric acid:= Mass / Mr= 126g / HNO3= 126g / (1) + (14) + (16 x 3)= 126g / (1) + (14) + (48)= 126g / 63= 2 molsodium nitrate:nitric acidMoles ratio 1 : 1Actual moles in reaction 2 : 2Mass of sodium nitrate:= Moles x Mr= 2 x NaNO3= 2 x (23) + (14) + (16 x 3)= 2 x (23) + (14) + (48)= 2 x 85= 170gMass of sodium nitrate needed = 170g
How many moles of silver will be generated if 1.30 moles of zinc is placed into the silver nitrate solution?
2.60 moles
How many moles of NaN03 are in a 2.70 m solution made with 3250g of H2O?
2.70 m = 2.70 moles/kg solvent3250 g solvent = 3.25 kg solvent2.70 moles/kg x 3.25 kg = 8.775 moles NaNO3 = 8.78 moles NaNO3 (3 significant figures)
How many moles of solute particles are produced by adding one mole of sodium nitrate to water?
two moles
