the percentage of water in an unknown hydrated sal...
the percentage of water in an unknown hydrated salt is to be determined by weighing a sample of the salt, heating it to drive off the water, cooling to room temperature, and re-weighing. which procedural mistake would result in determining a percentage of water that is too low?(a) i only(b) ii only(c) both i and ii(d) neither i nor ii
i only
i just pour in the amount of water for the cup size i'm using. then, even if i have the setting on full carafe, it will only use the amount of water i put in there. 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 4 parts water to 1 part sugar--ex. 1 cup water and 1/4 cup sugar. 
mix together and just bring to a boil. 
remove from heat.
let cool to room temp.
use a feeder that has red on it--do not add coloring to the nectar. hey there! as answered previously:-
crystallization is the process of slow precipitation of solid
crystals of a substance from a concentrated homogeneous solution it
is basically a separation method in which a substance is
transferred from the liquid (i e solution) phase to the solid state
the substance is thus obtained in pure crystalline form and has no
impurity it is used by chemists for the purification of solid
substances
following is an example of using crystallization to obtain pure
copper sulphate from an impure sample
1) take a small amount (5 g) of an impure sample of copper
sulphate in a china dish
2) dissolve the sample in 10 ml of water (we basically have to
use minimum amount of water in order to prepare a saturated
solution)
3) then, filter it to remove impurities
4) evaporate the filtrate on a water bath till small crystals
are formed indicating that a concentrated solution has been
obtained
5) after this, cover the solution with a filter paper and leave
it undisturbed at room temperature for the rest of the day
![](https://img-nm%20mnimgs%20com/img/study_content/lp/1/9/8/114/318/760/662/5-6-08_sonali_9_chemistry_9%202%202%202%201_gsx_sg_ss_html_3c9d32fa%20png)
it will be observed that crystals of pure copper sulphate are
obtained and the impurities are left behind in the solution (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. 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 (a) the candidate was studying the effect/importance of sunlight on the rate of photosynthesis.explanation: the candidate had covered one part of the leaf to study the rate of photosynthesis in the absence of sunlight. the candidate would compare it with the rate of photosynthesis on the uncovered parts of the leaf and thus, determine the role of sunlight during photosynthesis.(b) yes, the uncovered portion of the leaf of the potted plant is used as the control here.explanation: the control is used to compare the results which will be obtained after the experiment with the covered portion of the leaf to obtain conclusions.(c) plant was kept in the dark before the experiment, to remove the starch present in the leaf, to the storage organs.explanation: destarching is done by placing the plant for 24-48 hours in darkness before starting the experiment. this is done to make sure that all starch has been formed under experimental conditions only.(d) 1) first, the candidate would have dipped the prepared leaf in boiling water to kill the leaf cells.2) then he would have boiled the leaf in methylated spirit to remove the chlorophyll.3) the leaf would have then been placed in hot water to soften it.4) he would have placed it in a dish and poured iodine solution. the presence of starch is confirmed in regions that turn bluish-black in colour.the uncovered portion of the leaf would have turned bluish black and the cover region would have turned brown in colour.explanation: the uncovered portion of the leaf would have turned bluish black as starch would have formed in those regions. sunlight activates the chlorophyll, which in turn catalyzes the photosynthesis reaction. in the covered portion, on which sunlight could not have penetrated, the region turns brown due to the absence of chlorophyll. this implies that photosynthesis would not have taken place on the covered portion of the leaf. 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 ?solution:
step 1 of 2:
the intermediate pressure in a compressor is given by . the relative uncertainty of and the relative uncertainty of is 2% . we have to find the relative uncertainty in .


\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_




step 2 of 2:


the intermediate pressure in a compressor is = ((


it is given that the relative uncertainty in , = 0.05
 
 relative uncertainty in , = 0.02


so the relative uncertainty in the intermediate presser 




 = 


 = 
 
 = 0.0269


 therefore the relative uncertainty in the intermediate presser is 0.0269 mpa.
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ brief answer: consideration detailed answer: hello and thank you for sending the question. dehydration can lead to some degree of kidney dysfunction. this may be what they referred to as bruising. when the kid he does not have adequate perfusion of blood and fluid it can affect the overall...