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13 days ago

14

Write the balanced equation for each of the following changes and identify whether heat is a product or needed to start the reac

tion (do not include 'heat' in your balanced equation, simply identify whether it is a product or reactant):
(a) the sublimation of dry ice [conversion of CO2(s) directly to CO2(g)] Heat is a product and is given off by the reaction. Heat is a reactant and is needed to start the reaction.
(b) the reaction of 1 mol of sulfur dioxide with oxygen. Heat is a reactant and is needed to start the reaction. Heat is a product and is given off by the reaction.

1 answer:

Tems11 [2.4K]13 days ago

5 0

Clarification:

(a) sublimation of dry ice [the transition of CO2(s) straight to CO2(g)]

The balanced chemical equation is as follows;

CO2(s) --> CO2(g)

This process absorbs heat, which means; Heat serves as a reactant and is essential for initiating the reaction.

(b) the interaction of 1 mole of sulfur dioxide with oxygen.

The corresponding balanced equation is;

2SO2 + O2 --> 2SO3

This process releases heat, thus; Heat is a product and is emitted during the reaction.

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Approximately 220 million tires are discarded in the U.S. each year. These tires present a disposal problem because they take up

lions [2653]

Answer:

A total of 2667 tires are required to satisfy the annual power needs of ten homes.

Explanation:

According to the Second Law of Thermodynamics, not all energy produced when tires are incinerated can be effectively used due to losses associated with finite temperature differences. The energy obtainable from a tire when burned, measured in kilowatt-hours ( $E_{out}$ ), can be calculated using the efficiency definition:

$E_{out} = \eta \cdot E_{in}$

Where:

$\eta$ - Efficiency, which is dimensionless.

$E_{in}$ - Energy released from burning, measured in kilowatt-hours.

Taking into account $\eta = 0.5$ and $E_{in} = 75\,kWh$ , the yearly energy yield from a tire amounts to:

$E_{out} = 0.5\cdot (75\,kWh)$

$E_{out} = 37.5\,kWh$

Thus, the number of tires necessary to meet the electricity demand of ten homes for one year is:

$n = \frac{(10\,homes)\cdot \left(10000\,\frac{kWh}{home} \right)}{37.5\,\frac{kWh}{tire} }$

$n = 2666.667\,tires$

A total of 2667 tires are necessary to satisfy the annual power needs of ten homes.

8 0

28 days ago

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

Tems11 [2403]

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

8 days ago

The density of an alcohol is 0.788 g/mL. What volume in microliters, μL, will correspond to a mass of 20.500 mg?

lions [2653]

Answer:

B. 26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

$V=\frac{m}{\rho} =\frac{25.000mg}{0.788g/mL}*\frac{1g}{1000mg}*\frac{1000\mu L}{1mL} \\ \\V=26.0\mu L$

Thus, the solution is B. 26.0 μL.

Best regards.

6 0

13 days ago

An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%)

Tems11 [2403]

Answer:

The correct choice for your inquiry is option A, Argon.

Explanation:

Isotope Atomic mass Percent (%)

1 35.9675 0.337

2 37.9627 0.063

3 39.9624 99.6

To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

Theoretical Atomic mass

a) Ar 39.95

b) K 39.10

c) Cl 35.45

d) Ca 40.08

5 0

26 days ago

0.036549 round to three sig figs

Alekssandra [2719]

3 first significant figure
6 second significant figure
5 third significant figure
4 cannot exceed 5, so retain 5 instead of increasing it to 6

0.0365

6 0

28 days ago

Read 2 more answers