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I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
Without knowing the temperature, we'll assume 1 kilogram = 1 litre of water. Therefore if it is 5.1 mole per litre, it is 5.1 molar.
you simply divide the gram amount of sugar by the gram amount of water, and multiply by 100. So the mass percent is (2.45/200)x100=1.225%
Theoretically, the magnesium ion undergoes a tiny amount of hydrolysis. But this amount is so small that it realistically has no effect on the solution. So, in reality, it is neutral.
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.
I assume you mean 0.5 molar, but I'll go with what I'm given. If you have 500 molar of barium chloride, you will have 1000 molar of chloride ions as there are 2 chlorides for every 1 molecule of barium chloride. Having the 100ml there is irrelevant as you are talking about concentration and didn't ask for moles.
Pretty concentrated! ( I assume you mean 2.3 Liters )First get the moles of sodium chloride.456 grams NaCl (1 mole NaCl/58.44 grams)= 7.803 moles NaCl==============Now,Molarity = moles of solute/Liters of solutionMolarity = 7.803 moles NaCl/2.3 Liters= 3.4 M NaCl solution----------------------------
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
100 Liters? I will assume as much. Molarity = moles of solute/Liters of solution Molarity = 0.10 mole HCl/100.0 Liters = 0.001 M HCl -------------------------now, to find pH - log(0.001 M HCl) = 3 pH -----------------so, your acid is of 3 pH, which is to be expected at the volume od solution
That depends upon the concentration of the solution. If it is a 1 molar solution, then 2 ml contain .002 moles.
The standard enthalpy for the change in solution is: +3,87 kJ/mol at 25 0C
Assume purity unless otherwise stated. Find out the density and apply it to 1 litre to find out how many grams a litre will weigh. divide this answer by the molecular weight to get the number of moles in a litre
No, this cannot be answered. This is explained below:Please remember thatdensity is in g / mL (gram per millilitre)and molar mass is in g / mol (grams per mole)and molarity is in mol / mL (moles per millilitre)Without molar mass (in g/mol) molarity cannot be calculated from just knowing it is an triglyceride with density 0.91 g/mL (which is of no significance for this question, many trig's have a comparable density).
Without knowing the temperature, we'll assume 1 kilogram = 1 litre of water. Therefore if it is 5.1 mole per litre, it is 5.1 molar.
Ventolin contain an isotonic solution of salt; I assume that pharmacists have thought better. Sodium chloride has also an antibacterial effect.
Using this ideal gas equation PV=nRT. The question provided the following: P=2.61 atm V=2.14L R=0.0821 ( this is standard for any ideal gas) T= 28 C+273K= 301 n= unknown Since there is no volume change, so the volume is constant from the beginning of the reaction to the the end of the reaction. The only key information we are missing is the actual number of moles (n). We must rearrange the ideal gas equation to solve for moles. n=PV/RT n=(2.61)(2.14)/(.0821)(301) n=.2260 The question wants us to find the Molarity of the acid solution. We were given that HCL completely dissolved in 668mL of water. Molarity moles/Liters. 668mL=.668L M=.2260mol/.668L so Molarity of the acidic solution is .338.
Since copper (I) chloride has only limited solubility in water I will assume you mean copper (II) chloride, CuCl2. Then the ions will be Cu2+ and Cl-.