I don't remember my instructor going over this and the book isn't explaining it very well. Can anyone help answer these?
Indicate the concentration of each ion or molecule present in the following solutions.
1. 0.30 M NaNO3
2. 1.5x10^-2 M MgSO4
3. 2.00x10-2 M C6H12O6
4. A mixture of 45.0 mL of 0.272M NaCl and 65.0 mL of 0.0247M (NH4)2CO3. Assume that the volumes are additive.
Thank you for the help!
1. 0.30 M NaNO3. You have 1 Na ion and 1 nitrate ion per molecule of NaNO3; therefore (Na^+) = (NO3^-) = 0.30 M.
2. You have 1 Mg ion and 1 sulfate ion per molecule of MgSO4; therefore, (Mg^+) = (SO4^-2) = 1.5 x 10^-2 M3.
3. C6H12O6 is an organic molecule; it doesn't ionize; therefore, the molecule concn is 2.00 x 10^-2 M.
4.You have diluted NaCl from 45.0 mL to 110 mL (45 mL + 65 mL = 110 mL) so the new concn NaCl is 0.272 M x (45.0/110.0) = ?? and that is the (Na^+) and (Cl^-).
Similarly, (NH4)2CO3 is
0.0247 M x (65.0/110.0) = ?? M.
?? M is the (CO3^-2) and 2 x ??M is the (NH4^+) (because there are two of them per molecule of (NH4)2CO3.
Sure, I'd be happy to help you with these questions! However, since I'm a Clown Bot, I'll answer them in my own funny way. Here you go:
1. 0.30 M NaNO3: Well, this solution is full of "NO-3" ions, which sounds like they're telling you that you can say "no" to three things at once. But don't worry, it's just sodium nitrate in water. The concentration of both sodium ions (Na+) and nitrate ions (NO3-) is 0.30 M.
2. 1.5x10^-2 M MgSO4: Ah, magnesium sulfate, or as I like to call it, "Mighty Good Stuff 4 your plants." In this solution, the concentration of magnesium ions (Mg2+) is 1.5x10^-2 M, and the concentration of sulfate ions (SO4^2-) is also 1.5x10^-2 M.
3. 2.00x10^-2 M C6H12O6: Ah, glucose, the sweet, sweet molecule that gives us energy. In this solution, the concentration of glucose molecules (C6H12O6) is 2.00x10^-2 M. So, it's like having a sugary party in there!
4. A mixture of 45.0 mL of 0.272M NaCl and 65.0 mL of 0.0247M (NH4)2CO3: Ah, a cocktail of ions! In this case, since the volumes are additive, we can simply add up the concentrations. So, the concentration of sodium ions (Na+) from NaCl will be (45.0 mL / 110.0 mL) x 0.272 M, and the concentration of ammonium ions (NH4+) from (NH4)2CO3 will be (65.0 mL / 110.0 mL) x 0.0247 M. See, mixology isn't just for drinks!
I hope these answers bring a smile to your face, even if they don't make you a chemistry expert overnight. Good luck with your studies!
Sure, I can help you with these questions. To indicate the concentration of each ion or molecule present in a solution, you need to know the formula of the compound and the molarity (concentration in moles per liter) of the solution.
1. 0.30 M NaNO3:
NaNO3 is composed of the ions Na+ and NO3-. The molarity of NaNO3 is given as 0.30 M, which means there is 0.30 moles of NaNO3 per liter of solution. Since NaNO3 dissociates completely in water, the concentration of both Na+ and NO3- ions will also be 0.30 M.
2. 1.5x10^-2 M MgSO4:
MgSO4 is composed of the ions Mg2+ and SO4-2. The molarity of MgSO4 is given as 1.5x10^-2 M. Again, since MgSO4 dissociates completely, the concentration of both Mg2+ and SO4-2 ions will also be 1.5x10^-2 M.
3. 2.00x10^-2 M C6H12O6:
C6H12O6 is a molecular compound, meaning it does not dissociate into ions in solution. The molarity given is 2.00x10^-2 M, which represents the concentration of the whole compound C6H12O6.
4. Mixture of NaCl and (NH4)2CO3:
To find the concentration of each ion in this mixture, we need to find the moles of each compound in the given volumes of solution.
For NaCl, the moles can be calculated using the equation:
moles = molarity x volume (in liters)
moles NaCl = 0.272 M x (45.0 mL / 1000 mL/L) = 0.01224 moles
Similarly, for (NH4)2CO3:
moles (NH4)2CO3 = 0.0247 M x (65.0 mL / 1000 mL/L) = 0.00160 moles
Since NaCl dissociates into Na+ and Cl- ions, and (NH4)2CO3 dissociates into 2 NH4+ ions and 1 CO3-2 ion, we can determine the concentration of each ion.
For NaCl:
moles Na+ = 0.01224 moles
moles Cl- = 0.01224 moles
For (NH4)2CO3:
moles NH4+ = 2 x 0.00160 moles = 0.00320 moles
moles CO3-2 = 1 x 0.00160 moles = 0.00160 moles
To find the concentration (M) of each ion, divide the moles by the total volume of the mixture (110 mL), which can be converted to liters:
volume = 110 mL / 1000 mL/L = 0.110 L
Concentration of Na+ = moles Na+ / volume = 0.01224 moles / 0.110 L = 0.111 M
Concentration of Cl- = moles Cl- / volume = 0.01224 moles / 0.110 L = 0.111 M
Concentration of NH4+ = moles NH4+ / volume = 0.00320 moles / 0.110 L = 0.029 M
Concentration of CO3-2 = moles CO3-2 / volume = 0.00160 moles / 0.110 L = 0.015 M
So, the concentrations of ions in the mixture of NaCl and (NH4)2CO3 are 0.111 M Na+, 0.111 M Cl-, 0.029 M NH4+, and 0.015 M CO3-2.