which of the following aqueous solutions are good ...
which of the following aqueous solutions are good buffer systems? and why?a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.b. 0.28 m hydroiodic acid + 0.18 m sodium iodide.c. 0.30 m ammonia + 0.37 m ammonium nitrate.d. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid.e. 0.38 m calcium bromide + 0.30 m barium bromide.
we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate
a) ans. concentration = no. of moles of solute * 1000/ vol. in m;

no. of moles of solute = concentration * vol / 1000 = 1.4 * 270 / 1000 = 0.378 = 0.378 mol.



b)

molality = 1.65

molality = moles of solute / mass of solvent in kg

1.65 = moles of solute / 5.5 x 10^-5 kg

moles of solute = 9.075 x 10^-5

c)

mass of sucrose = 75 x 1.4 / 100 = 1.05g

moles = 1.05 / 342 = 3.07 x 10^-3 use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 reno, please don't answer organic chemistry questions if you don't know the subject matter.
 
mixing acetic acid with sodium bicarbonate doesn't yield "gaseous water." also, you didn't actually answer the question and decided instead to list the first thing you found in wikipedia about a dangerous reaction that has nothing to do with the fact that acetic acid and hcl react the same way because they are both stronger acids than the conjugate acid of bicarbonate. it is best to see doc for proper evaluation and treatment. there are otc med and rx med but work differently for different pts. one may respond better to you may not respond good with other pts.doctor can help you to find the causes of your chronic acidity such as stress, diet etc.. see more... AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v Tin(II) chloride can dissolve in less than its own mass of water without apparent decomposition, but as the solution is diluted, hydrolysis occurs to form an insoluble basic salt:

 

Therefore, if clear solutions of tin(II) chloride are to be used, it must be dissolved in hydrochloric acid (typically of the same or greater molarity as the stannous chloride) to maintain the equilibrium towards the left-hand side (using Le Chatelier's principle).

**Sources:** https://en.m.wikipedia.org/wiki/Tin(II)\_chloride images photos vector graphics illustrations videos… color . transparent black and white. related images: bathroom sink faucet plumbing water 168 free images of bathroom sink. 55 75 5. bathroom sink faucet. 55 71 4. bathroom clean faucet. 114 139 19. bathroom luxury. 128 139 16. bathroom… 168 free images of bathroom sink everyone responds differently so its hard to tell. its really a trial and error thing to find the opioid that works best for you at the moment. see more... (i) adsorption is a surface phenomenon that causes the accumulation of molecules of a substance at the surface of a solid or liquid rather than in the bulk. in adsorption, the substance gets concentrated at the surface only. it does not penetrate through the surface to the bulk of the solid or liquid. for example, when a chalk stick is dipped into an ink solution, only its surface becomes coloured.


on the other hand, the process of absorption is a bulk phenomenon. in absorption, the absorbed substance gets uniformly distributed throughout the bulk of the solid or liquid. for example, when a sponge is dipped in water the whole sponge gets wets.


(ii)according to the schulze-hardy rule, the effectiveness of the salt causing flocculation depends on the charge on the ion of opposite sign to the charge on the sol particles. the greater the magnitude of the opposite charge, the higher the ability of a salt to coagulate the sol.  
thus, trivalent salt alcl3 is more effective in causing the coagulation of a negatively charged sol than divalent salt mgcl2.


(iii)out of sulphur sol and proteins, sulphur sol forms multimolecular colloids. sulphur sol consists of particles containing a thousand or more s8 molecules. on the other hand, proteins form macromolecular colloids.