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igomit [66]
3 years ago
9

In a recrystallization (a technique that you will encounter later in the semester), a solid is dissolved in a solvent and later

the solvent is removed. if a mp of the sample is taken while the sample is still moist with solvent, what effect would that have on the mp of the sample?
Chemistry
1 answer:
grandymaker [24]3 years ago
7 0

In a recrystallization (a technique that you will encounter later in the semester), a solid is dissolved in a solvent and later the solvent is removed. if a mp of the sample is taken while the sample is still moist with solvent then melting point will not come out to be accurate, as solvent is still present in the compound and that results in the variation of the melting point as the residual solvent will serves as impurities , and due to the presence of these impurities, melting point vary and does not come to be accurate .

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The volume of the drum after chilled quickly is 3.6 L.

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
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Answer : The enthalpy of neutralization is, 56.012 kJ/mole

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First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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3 years ago
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