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bija089 [108]
3 years ago
8

A 20.00 ml sample of a solution of sr(oh)2 is titrated to the equivalence point with 40.03 ml of 0.1159 n hcl. what is the molar

ity of th sr(oh)2 solution?
Chemistry
1 answer:
goblinko [34]3 years ago
3 0
The   molarity  of Sr(OH)2  solution is  =  0.1159 M

    calculation
write the equation  for reaction
that is,  Sr(OH)2 +2HCl→ SrCl2 + 2 H2O

then finds the mole  of HCl used

moles = molarity x volume 
=40.03 x0.1159 =  4.639 moles

by  use of mole ratio between Sr(OH)2 to  HCL which is 1 :2  the moles of Sr(OH)2  is therefore =  4.639  x1/2 = 2.312  moles

molarity  of  Sr(OH)2  is =  moles  / volume

=2.312 /20 =0.1159 M
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3 years ago
When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio
arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>

<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>

<em>m is molality of the solution</em>

<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>

<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>

5 0
2 years ago
what is the half life isotope if after 21 days you have 31.25 g remaining from a 250.0 g starting sample size?
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each arrow equals one half life, so 3 half lives


14 days/ 3 half lives =


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3 0
3 years ago
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