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.
Unfortunately this question is a complicated mathematical equation that can not be completed in 750 characters. There is a complex equation, where the user would in put the volume of the ions and the solution and calculate the solution in that manner.
Concentrated solution is a solution that contains a large amount of solute relative to the amount that could dissolve.
it is very easy to prepare working solution from a stock solution we use the formula for this purpose which is: C1V1 = C2V2 C1 is the concentration of the stock solution V1 required volume from the stock solution C2 concentration of the working solution V2 volume of the working solution
density = mass/volume mass = density x volume volume = mass/density
Vital Capacity (VC) is the maximum amount of air that can be exhaled after a maximum inhalation. But it differs from one person to another. For a normal sized male that would be 4600 ml.
Unfortunately this question is a complicated mathematical equation that can not be completed in 750 characters. There is a complex equation, where the user would in put the volume of the ions and the solution and calculate the solution in that manner.
It may trapped in your sample, if it's a polymer for example. Or it may reacted with a substance in the solution.
Your question does not make sense, an almost infinite amount of solution could be prepared if desired
The concentration of a solution is basically how strong the solution is.
The maximum volume for intradermal injection is 0.1 ml and the maximum volume for subcutaneous injection is 2ml.
The concentration of a solution is typically given in terms of the volume of solution, in liters.
Per cent by volume means mls of solute per 100 mls solution. So you need to know the volume of the solute and the total volume of the solution. Divide volume of solute by volume of solution and multiply by 100 to get per cent by volume.
20 volume is 6% solution. To make it 3% solution just add same volume of water to the original 6% solution and you have double volume of 3% solution.
If the concentration of alcohol and water solution is 25 percent alcohol by volume, the volume of alcohol in a 200 solution is 50.
It depends if the mass of solute is given volume a solution, or mass/volume.
Le Volume Était Au Maximum was created in 2000.
Gas