Answer to Question #205673 in General Chemistry for Bill




What volume in milliliters of a 0.141

MNaOH solution is required to reach the equivalence point in the complete titration of a 9.00

mL sample of 0.122 MH2SO4?
V(NaOH) = 2×9×0.122/0.141 = 15.57 mL
recall that at the equivalence point of a titration:[readmore]moles acid=moles basealso, recall that moles = molarity × volume. solving for molarity of naoh:0.568 g 1 mol khp204 g1 mol oh-1 mol khp1 mol naoh1 mol oh-136.78 ml1 ml1x10-3 l = 0.0757 mollor mtherefore, the molarity of naoh needed is 0.0757 m for water, h2o1 mol h2o = 18.0 g h2o1 mole h2o = 6.022x1023 h2o molecules[readmore]the number of h2o molecules in a 9.00-g sample of water:9.00 g h2o×1 mol h2o18.0 g h2o×6.022×1023 h2o molecules1 mol h2o= 3.01x1023 h2o moleculesthe number of h2o molecules in a 9.00-g sample of water is 3.01x1023 h2o molecules. 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 setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m Correct option is A) 9.378 m2


![](https://lh3.googleusercontent.com/-UueuqyijXPQ/YLtK0v0m9CI/AAAAAAAABro/xFcfQOQAOx8jHAtpqiORvW_8DRFX1yK3ACLcBGAsYHQ/image.png)


result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits) 0.10m hc2h3o2 and 0.10m lic2h3o2 mass (molar mass) → mol (avogadro's number) → molecules[readmore]3.51×104 g ×1 mol163.94 g×6.022×1023 molecules1 mol = 1.29x1026 moleculesthere are 1.29x1026 molecules in the sample 讓戰將來回答~~

圖片參考：http://www.molecularstation.com/images/western-blot.gif

 0.05%tbst
1. pour 3 l of purified water into a 4-l beaker and begin stirring.
2. measure 400ml 10x tbs in a 500-ml graduated cylinder and add to water.
3. add bsa and 2ml tween 20 to the beaker. rinse all of the tween 20 from the pipette tip into the solution.
4. transfer solution to a graduated cylinder and adjust to final volume 4l with purified water.
5. return solution to beaker and stir until thoroughly mixed.
以上是以 4l 為配置體積, 若你用不到那麼多可依比例減少配置體積, 而 2%bsa 的配法則依此要領, 將你們實驗室中用的 bsa 加入即可, 因為不知道你們是用 powder 還是 liquid 的 bsa~

圖片參考：http://www1.qiagen.com/products/genesilencing/allstarrnaicontrols/images/western.jpg

希望以上有幫到您~~ radius = 1.75 m
height (h) = 12 m
volume = 
= ×  × 1.75 × 1.75 × 12
= 38.5 m3
since, 1 m3 = 1 kiloliter
capacity of the pit = (38.5 × 1)
= 38.5 kilolitres 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