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Mila [183]
3 years ago
6

The Lewis dot model of a molecule is shown. Based on the model, which of the following is true? (5 points)

Chemistry
1 answer:
vichka [17]3 years ago
6 0

Answer:

B i took the test

Explanation:

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At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

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

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
A gas occupies 12.3 liters at a temperature of 40.0k what is the volume when the temperature is increased to 60.0k
blagie [28]

Answer:

18.45L

Explanation:

V1 = 12.3L

T1 = 40K

T2 = 60K

V2 = ?

This question involves the use of Charles law which states that the volume of a fixed mass of gas is directly proportional to its temperature provided that pressure remains constant.

Mathematically,

V = kT k = V/T

V1 / T1 = V2 / T2 = V3 / T3 =.......=Vn / Tn

V1 / T1 = V2 / T2

V2 = (V1 × T2) / T1

V2 = (12.3 × 60) / 40

V2 = 18.45L

The final volume of the gas is 18.45L

6 0
3 years ago
21. List 3 indicators of a chemical change.
AfilCa [17]

21. change in temperature, change in color, or noticeable odor (after reaction has begun)

22. Picture...  Wasn't so sure about this since no charge was given, so I suspected it has a charge of 2. If you have more information on this I would be glad to help, and I can't tell what is happening with the calculations given. It would be helpful if you can add spaces or put a line break.

23. An ion is an atom or molecule with an electric charge. It is formed when atoms lose or gain electrons. There are two types of ions: cations (net positive charge) and anion (net negative charge).

24. Picture

25. a) dinitrogen tetrahydride

     b) phosphorus triiodide

     c) lithium hydroxide

5 0
3 years ago
Read 2 more answers
Given that the frequency of a wave is 9.12x10^-12 Hz, what must be the
polet [3.4K]

Answer:

The wavelength of wave is 0.33 ×10²⁰ m.

Explanation:

Given data:

Frequency of wave = 9.12×10⁻¹² Hz

Wavelength of wave = ?

Solution:

Formula:

Speed of light = wavelength × frequency

c = λ × f

λ = c/f

This formula shows that both are inversely related to each other.

The speed of light is 3×10⁸ m/s

Frequency is taken in Hz.

It is the number of oscillations, wave of light make in one second.

Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.

Now we will put the values in formula.

λ = 3×10⁸ m/s  / 9.12×10⁻¹² Hz

Hz = s⁻¹

λ = 0.33 ×10²⁰ m  

The wavelength of radiation is 0.33 ×10²⁰ m  .

5 0
4 years ago
The enthalpy of formation of CO2 at 25oC is -393.51 kJ/mol. What is the enthalpy of formation at 500 oC?
Kisachek [45]

Answer:

-375.9_KJ/(mol)

Explanation:

H(T2 ) ≈ H(T1)+CPΔT

Specific heat of Carbon is 0.71 J/g K.

At 283.15 the heat capacity is 37.12 J/(mol*K)

Kirchhoff's law

H(T2 ) ≈ H(T1)+CPΔT

Where

H(T1) and H(T2 ) are the heat of formation of CO2 at temperatures T1 and T2

CP is the heat capacity

Thus we have and ΔT is the temperature change

H(T2 ) ≈ -393.51×10^3+CP×(500-25)

= -393.51×10^3+37.12×(500-25)

= -375878 J/(mol)

= -375.9KJ/(mol)

6 0
3 years ago
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