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tatuchka [14]
3 years ago
5

The energy released in a nuclear reaction comes from

Chemistry
2 answers:
slavikrds [6]3 years ago
6 0

The energy is released from the "conserved" energy of an atom

riadik2000 [5.3K]3 years ago
6 0

it comes from some of the mass deteriorating into photons

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Is carbon monoxide or oxygen more ideal gas
Marianna [84]

Answer: Oxygen

Explanation:

7 0
4 years ago
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Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c
Alla [95]

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)=\frac{W}{nR}

(T₂-T₁)= \frac{2.40 X 10^{3} }{6 X 8.314}

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.


6 0
3 years ago
What is the most likely pH of apple juice?
asambeis [7]
7...................
8 0
3 years ago
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data: ΔH∘rxn 180.5 kJ/mol ΔS∘
nasty-shy [4]

Answer:

7.28 × 10³ K

Explanation:

Let's consider the following reaction.

N₂(g) + O₂(g) → 2 NO(g)

The reaction is spontaneous when the standard Gibbs free energy (ΔG°) is negative. ΔG° is related to the standard enthalpy of the reaction (ΔH°) and the standard entropy of the reaction (ΔS°) through the following expression.

ΔG° = ΔH° - T . ΔS°

If ΔG° < 0,

ΔH° - T . ΔS° < 0

ΔH° < T . ΔS°

T > ΔH°/ΔS° = (180.5 × 10³ J/mol)/(24.8 J/mol.K) = 7.28 × 10³ K

The reaction is spontaneous above 7.28 × 10³ K.

7 0
4 years ago
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
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