what is the solubility product, ksp of mg(oh)2 if ...
what is the solubility product, ksp of mg(oh)2 if its solubility in water is 1.6 x 10 -4 mol•l-1?

a) 1.6 x 10-11

b) 2.6 x 10-8

c) 3.2 x 10-4

d) 4.1 x 10-12
1.6 x 10-11
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 Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm 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. 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](#) 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 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 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** on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2- (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