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2 years ago

titration 25.0 ml of 3.4 x 10^ -3 M of Ba(OH) Neutralize 16.6 mL of HCL solution. What is the molarity of HCL solution?

1 answer:

8 0

What we are give: Concentration of base (CB) = 3.4 × $10^{-3}$
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve.
CA= $\frac{CBVB}{VA}$

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Answer:

$T_f$ = 25.05°C

Explanation:

Given:

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molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

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since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

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symbol: ¹¹⁵₅₀Sn .

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