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Maru [420]
3 years ago
7

Suppose a student completes an experiment with an average value of 2.6 mL and a calculated standard deviation of 0.80 mL. What i

s the minimum value within a 1 SD range of the average
Chemistry
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

Explanation:

1 standard deviation range will look like: [average - SD, average + SD]

Here the miximum value will be: average - SD = 2.6 - 0.80 = 1.80 ml

and the maximum value will be: average + SD = 2.6 + 0.80 = 3.40ml

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no it is not a complete model

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3 years ago
Salt dissolved in water is a solution, therefore _____.
saveliy_v [14]

Answer: Correct options are as follows.

  • salt is not chemically bonded to water.
  • salt and water retain their own chemical properties.

Explanation:

When salt is dissolved in water then it means that it is a physical change as salt has completely dissociated into ions but they are not chemically combined to the water molecules.

As a result, both salt and water will retain their chemical properties.

For example, NaCl when dissolved in water will dissociate as follows.

           NaCl \rightarrow Na^{+} + Cl^{-}

Only the particles of salt have evenly distributed in water.

And, when a components of a salt chemically combine with another substance then it will form a new compound.

Therefore, we can conclude that salt dissolved in water is a solution, therefore:

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4 0
3 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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Answer:

I think option A..

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