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
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
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
NaNO3 is sodium nitrate.
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.
A reaction of ammonia and water make ammonium hydroxide.I think ammonium hydroxide is a weak base. it has most of ammonia molecule because ammonia has a property of nucleophilic and same thing is done by the ammonium hydroxide.
The molarity is 5,55.
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
NaNO3
Make sure that the equation is balanced. NaOH + HNO3 > > NaNO3 + H2O ( all one to one ) 15.0 grams NaOH (1mol NaOH=39.998g) (1mol HNO3/1mol NaOH) = 0.375 moles of HNO3 Molarity = mols solute/volume solution 2.00M HNO3 = 0.375 mols/X volume = 0.188 Liters or, 188 milliliters.
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.
Sodium nitrate is neutral in water solution..
NaNO3
Sodium nitrate (NaNO3) is very soluble in water.
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
it is O
A reaction doesn't exist; the solution contain ions of calcium, iodine, sodium and nitrate.
NaCI + AgNO3 >>>>> NaNO3 + AgCI (white ppt.)