a student weighs out 0.568 g of khp (molar mass = ...
a student weighs out 0.568 g of khp (molar mass = 204 g/mol) and titrates to the equivalence point with 36.78 ml of a stock naoh solution. what is the concentration of the stock naoh solution? khp is an acid with one acidic protona) 3.15 mb) 0.0757 mc) 0.100 md) 0.943 me) none of these
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
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 V(NaOH) = 2×9×0.122/0.141 = 15.57 mL 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** 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 given radius of cone (r) = 3.5/2 m height of cone (h) = 12 m find out we have to find the capacity in kiloleters solution volume of cone = (1/3) × πr2h on taking π=22/7 = (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 = (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2) = 38.5 m3 capacity of conical pit in kilo liters: 1000 m3 = 1 liter 38.5 m3 = 1000 × 38.5 liters = 38,500 liters = 38.5 kilo liters answer the capacity of the container is 38.5 kilo liters Correct answer is 26 mm. Mass of powder is constant. mi=mf  ⇒ ρiVi = ρfVf ρihi = ρf hf ⇒ hi = (ρf hf)/ρi On substitution, (1755×12)/ 810= 26 mm  


Correct answer is 26 mm.


Mass of powder is constant.


mi=mf ⇒ ρiVi= ρfVf


ρihi = ρf hf⇒ hi = (ρf hf)/ρi


On substitution, (1755×12)/ 810= 26 mm


 


[See less](#) (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters 1.5 kg = 1.5 × 1000 g = 1500 g∴ ratio of 1.5 kg to 2500 g= 1500 g : 2500 g = `1500/2500` = `3/5` = 3 : 5 answer: the molecular mass of an element is defined as the sum of the masses of the elements present in the molecule. molecular mass is obtained by multiplying the atomic mass of an element with the number of atoms in the molecule and then adding the masses of all the elements in the molecule. o2 is a diatomic molecule. the mass of an oxygen atom = 16 amu. ∴ mass of o2 molecule = 2 × 16 = 32 amu. mass of one molecule of oxygen → 32/ avogadro constant ( 6.023 × 1023) → 5.31 × 10-23g. 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