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1 month ago

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Write a hypothesis for Part II of the lab, which is about the type of material an object is made of, and its ability to absorb o

r release thermal energy. Use the format of "if . . . then . . . because . . .” and be sure to answer this part of the lesson question: "How does the type of material affect thermal energy transfer?”

2 answers:

castortr0y [3K]1 month ago

5 0

Answer:

Various materials will exhibit different temperature changes when subjected to an equal amount of thermal energy. This occurs because each material has its distinct specific heat.

Explanation:

that is the answer

Tems11 [2.7K]1 month ago

5 0

Answer:

For the second part of the lab, my hypothesis would be

Explanation:

If materials of equal volume are subjected to the same thermal energy, then they will transfer varying amounts of energy, given that each substance possesses a specific heat unique to it.

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A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

25 days ago

A chemist is studying the following reaction: NO + NO2 ⇌ N2O3. She places a mixture of NO and NO2 in a sealed container and meas

castortr0y [3046]

Answer:

The forward reaction will keep occurring until all NO or all NO₂ is consumed.

Clarification:

According to Le Châtelier's principle, when a system at equilibrium experiences a disturbance from an outside source, the system will adjust to counteract this disturbance and restore equilibrium.

Thus, removing the product (N₂O₃) from the system effectively lowers the product concentration, prompting the reaction to shift forward and generate additional product in order to alleviate the strain caused by the removal of N₂O₃.

Consequently, the reaction will proceed forward until all of either NO or NO₂ is depleted.

5 0

1 month ago

Select the word or phrase from the drop-down menu to describe ionic compounds. A formula unit represents the simplest ratio of e

VMariaS [2998]

Response:

CRYSTAL

A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE

1:1

Reasoning:

8 0

1 month ago

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Select the option that correctly completes this statement: If the reaction takes place at a constant volume in a thermally insul

KiRa [2933]

Answer:

The temperature of the gas rises.

Explanation:

This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.

The gas only receives internal energy from the heat transferred to it from the surroundings.

In this situation, the pressure also increases.

8 0

28 days ago

The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab

Anarel [2989]

Answer:

Please review the following responses

Explanation:

1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)

2) 100. mL of NaCl solution at 3.00 M (3 M)

3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)

4) The concentrations of beakers 1 and 5 are identical (1.5M)

Molar mass of NaCl = 23 + 36 = 59 g

For beaker number 3:

59 g -------------- 1 mol

19.5 g ------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For beaker number 4:

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For beaker number 5:

Molarity (M) = 0.450/0.3 = 1.5 M

4 0

1 month ago

Read 2 more answers