Help finding an unknown cation?
Suppose you were given a nitrate compound with an unknown cation to test. After combining it with many different solutions, you are unable to produce a precipitate. a) What are some possible cations that could give these results? Explain why. b) What method could you use to determine which of these cations was present in your salt? Explain how you would use this method to determine the identity of the unknown cation Thank you so much!!
AP chem lab help: unknown sulfate salt and have to find the mass %?
I'm a little confused by your question. I don't think that barrium sulfide (BaS) will be precipitating out of your solution. It's much more likely to be barrium sulfate (BaSO4). Assuming that you're looking for the % mass of sulfur by looking at how much BaSO4 you have, here's what to do: Take your mass of BaSO4 and convert to moles (by dividing by molar mass). Determine the number of moles of sulfate based on the number of moles of barium sulfate (1:1 mole ratio, so should be same #). Now, determine the number of moles of S from number of moles of SO4^2-. (1 mole of S in every mole of sulfate, so 1:1 mole ratio, so should be same # again) You now have the moles of sulfur in you barium sulfate salt. All of the sulfate in barium sulfate must have come from you original unknown sulfate salt. That also means that all of the moles of S must have come from that original moles of sulfate. Ok, so now you know the moles of S. Convert this to grams (multiply by molar mass). This is the mass of S in your original sample. Now divide the mass of S by the total sample mass and multiply by 100 to get your % mass of S in unknown sulfate salt.
How to find the % of water of unknown salt?
Mass of water in the unkown hydrate = Mass of test tube and unknown hydrate - Mass of test tube and anhydrous salt = 7.294 g - 6.748 g = 0.546 g Mass of the unknown hydrate = Mass of test tube and unknown hydrate - Mass of test tube = 7.294 g - 5.781 g = 1.513 g the % of water of unknown salt = Mass of water in the unkown hydrate/Mass of the unknown hydrate = 0.546 g/1.513 g = 36.087%
Unknown blue salt with anion SO4 2-. Finding Cation....(Gen Chem II Lab)?
I am in Gen Chem II lab studying general unknown salts. My salt was blue, like sky blue? I went through the anion process and found SO4 2-. Afterwards, I dissolved some of the salt in H2O. I mixed this solution with NH3. The solution turned dark blue like aqueous Copper (II). Do I have CuSO4? I was told I might of done the process wrong because I did not remove my anion...... and I was told I might of done the process wrong because I did not add HCl, Ta, NH3, centrifuge, decant, and etc... (a whole process of precipitating group II ions). However, I was with my lab professor and she said I should go ahead and use the new solution and just add NH3, but she's not the main lab professor.. so I just want to make sure I did this right. Thanks!
How to figure out delta H for an unknown salt?
If I have a calorimeter with a heat capacity of 40 J/degree C and I add 100 g of water to that calorimeter then I add 4 grams of an unknown salt to the water+calorimeter, How do I find out delta H of the unknown salt in J/g of salt? I also have a temperature change of -2 degrees C after the addition of the unknown salt. This is what I did, is it correct? delta H(calorimeter) = C(calorimeter) * delta T delta H(calorimeter) = 40 * -2 delta H(calorimeter) = -80 J delta H(reaction) = -delta H(calorimeter) delta H(reaction) = 80 J delta H(reaction) = 80 / 4 delta H(reaction) = 20 J/g of salt Thanks
Enthalpy of Unknown Salt?
Ok, We did a lab today in chem class. And I'm having some trouble. Purpose: Use Calorimetery to determine the enthalpy of soultion of an unkown salt and then use that value to identify the salt in the medical cold pack We used 10g of the unkown salt and we used 100ml of water My lab results were: Change in temprature of unkown soultion: 27C-21C= 6C Our teacher gave us this chart to check our answer: Salts in a Medical cold pack Ammonium Chloride .277 kj/g potassium nitrate .345kj/g sodium actate trihydrate .144kj/g potassium chloride .231 kj/g
Calculating Ka from pH of an unknown salt solution?
First calculate c (concentration) , c = Ca = n/V Then calculate [H3O+] , [H3O+] = 10^-pH then calculate Ka , Ka = [H3O+]^2 / Ca Calculate them using the values that you have: n = 0.1 M V = 15mL=0.015L pH = 6.07
The identity (enthalpy of solution) of unknown salt.?
The heat transfer of water or specific heat of water is: Specific heat of water, 4.186 J/(g•°C) Enthalpy of solution = speccifc heat of water * (change in temperature) * weight of water. Change in temperature is the change in temperature of the water after you add the unknown salt to the water. So, if you have 100 g of water and the temperature changes from 40 C to 50 C, the enthalpy of solution is: 4.186 * 100 * (50-40) = 4186 J
How can we know a salt is acidic or basic?
Look , salts are generally neutral but sometimes they can be acidic or basic .Now if we want to know if its acidic or basic , the most fundamental approach to it is look for the H or OH ions in the salt .Consider the possibility of salt without any OH ions . If you get a neutral salt (when you remove any OH ions ) then the salt you had was basic .Same is for acidic salt . Just replace the H ion .What happens actually is , in a neutralisation reaction the OH ions of base are unable to completely react with the H ions of an acid .This results in the formation of a salt with excess of H ions or OH ions and hence the salt is termed as Acidic or Basic respectively !Hope this helps.Thanks for asking !