I have a chemistry test tomorrow and I'm trying to study but none of these questions in the book make sense to me:(
Help!
1. Commericially available concentrated sulfuric acid is 95% H2SO4 by mass and has a density of 1.84g/mL. How many milliliters of this acid are needed to give 1.0 L of .1 M H2SO4?
2. 2.488g of an impure sample of Na3PO4 are dissolved in water and all of the PO43- ions are precipitated to give 1.796g of Ca3(PO4)2. What is the percentage of Na3PO4 in the original sample?
3. What is the molarity of a solution made by diluting 75.5 mL of .15M(molarity) of KOH to exactly 250 mL?
4. What volume of .10M HCL reacts completely with .45g Zn if the reaction products are ZnCl2 and H2?
please help=/
Thanks!
Can someone help please? I'm not asking for answers, just how to go about them=/
Idk lol
Sure, I can help you understand how to solve these chemistry problems. Let's go through each question step by step:
1. To solve this problem, you need to first determine the number of moles of H2SO4 needed to make 1.0 L of 0.1 M H2SO4. The molarity (M) is defined as moles of solute per liter of solution. Since the molarity is 0.1 M and the volume is 1.0 L, you have 0.1 moles of H2SO4.
Next, you need to calculate the mass of H2SO4 required to give 0.1 moles. Knowing that commercial sulfuric acid is 95% H2SO4 by mass, you can use the formula:
mass = moles x molar mass
The molar mass of H2SO4 is 98.09 g/mol. So, the mass of H2SO4 required is:
mass = 0.1 mol x 98.09 g/mol = 9.809 g
Finally, you need to find the volume of concentrated sulfuric acid needed to give this mass. The density of the acid is 1.84 g/mL. Using the formula:
volume = mass / density
volume = 9.809 g / 1.84 g/mL = 5.33 mL
Therefore, you would need approximately 5.33 mL of the concentrated sulfuric acid to prepare 1.0 L of 0.1 M H2SO4.
2. To find the percentage of Na3PO4 in the original sample, you need to calculate the mass of Na3PO4 in the precipitate and divide it by the mass of the original sample, then multiply by 100.
First, determine the number of moles of Ca3(PO4)2 in the precipitate. The molar mass of Ca3(PO4)2 is 310.18 g/mol. So the number of moles is:
moles = mass / molar mass = 1.796 g / 310.18 g/mol = 0.0058 mol
Since 2 moles of Na3PO4 are required to produce 1 mole of Ca3(PO4)2, the number of moles of Na3PO4 is twice that of Ca3(PO4)2:
moles Na3PO4 = 2 x 0.0058 mol = 0.0116 mol
Next, calculate the mass of Na3PO4:
mass Na3PO4 = moles Na3PO4 x molar mass = 0.0116 mol x 163.94 g/mol = 1.901 g
Finally, determine the percentage of Na3PO4 in the original sample:
percentage = (mass Na3PO4 / mass of original sample) x 100 = (1.901 g / 2.488 g) x 100 = 76.38%
Therefore, the percentage of Na3PO4 in the original sample is approximately 76.38%.
3. To find the molarity of the diluted KOH solution, you need to use the formula:
M1V1 = M2V2
where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
In this case, M1 is 0.15 M, V1 is 75.5 mL, M2 is the unknown, and V2 is 250 mL.
Using the formula, you can calculate the molarity of the diluted solution:
(0.15 M) x (75.5 mL) = M2 x (250 mL)
Solving for M2, you get:
M2 = (0.15 M x 75.5 mL) / 250 mL = 0.045 M
Therefore, the molarity of the solution after dilution is 0.045 M.
4. To determine the volume of 0.10 M HCl needed to react completely with 0.45 g of Zn, you need to use the stoichiometry of the reaction and the balanced chemical equation.
The balanced equation for the reaction between HCl and Zn is:
Zn + 2HCl -> ZnCl2 + H2
From the balanced equation, you can see that the mole ratio between Zn and HCl is 1:2. This means that 1 mole of Zn reacts with 2 moles of HCl.
First, calculate the number of moles of Zn:
moles of Zn = mass / molar mass = 0.45 g / 65.38 g/mol = 0.0069 mol
Since the mole ratio between Zn and HCl is 1:2, you need twice the number of moles of HCl:
moles of HCl = 2 x moles of Zn = 2 x 0.0069 mol = 0.0138 mol
Finally, calculate the volume of 0.10 M HCl required using the formula:
volume = moles / molarity
volume = 0.0138 mol / 0.10 M = 0.138 L = 138 mL
Therefore, you would need approximately 138 mL of 0.10 M HCl to react completely with 0.45 g of Zn.
I hope this helps you understand how to solve these chemistry questions! If you have any further questions, feel free to ask. Good luck with your test!