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3 years ago

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What do you predict would happen if you conducted this experiment using sulfur hexafluoride (sf6) which is much heavier than met

hane or butane?

1 answer:

Sophie [7]3 years ago

4 0

The volume of SF6 would be larger and this would affect the pressure in the flask .

Explanation:

If we conducted an experiment using same number of moles of sulfur hexafluoride (SF6) ,methane or butane, there will be no any effect on temperature, pressure and volume. But if we conducted an experiment using same weight of sulfur hexafluoride, methane or butane, there will be some effect on temperature, pressure and volume.

Hope that helps!

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All food chain in an ecosystem that are connected to each other forms a food web....

Therefore the answer is food web

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Making wire from a block of copper.<br><br> Physical Change<br> or<br> Chemical Change?

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Answer: B

Explanation:

protons are positive,

electrons are negative,

and neutrons are neutral.

the amount of electrons to protons is always the same in a balanced atom.

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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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Can someone convert 150 cm of copper (Cu) wire into mm of Cu wire?

Vedmedyk [2.9K]

1500 mm of copper wire are produced by converting 150 cm of copper wire.

The following is the centimeter to millimeter conversion factor: 1 cm is equivalent to 10 mm.

Consequently, 150 cm will equal 150 x 10 = 1500 mm.

The multi-step process of unit conversion involves multiplying or dividing by a numerical factor.
Unit conversion is the process of converting the measurement of a given amount between various units, often by multiplicative conversion factors that alter the value of the measured quantity without altering its effects.
Unit conversion is a multi-step procedure that involves adding, subtracting, multiplying, or dividing by a conversion factor.
The process may also require the selection of the correct number of significant digits, and rounding.

To learn more about unit conversion visit:

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