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Mademuasel [1]
4 years ago
12

Which has not been a major source of CFCs

Chemistry
2 answers:
Alik [6]4 years ago
6 0

Answer:

• televisions

ipn [44]4 years ago
4 0

Answer : Any natural sources of CFC's are not known only the major sources like aerosols, propellants, refrigerants,etc are known. So, if any natural sources are given then it cannot be called as a major source for emitting CFC into environment.

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How many molecules are in 1 mol of the chemical equation shown above
aivan3 [116]
There's 6.022×10^23 particles in 1 mole of anything

like there is 1000 grams in 1 kilogram of anything
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4 years ago
Hey I'm a newbie any help please with this
solniwko [45]
6 A
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9 B
10 A
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4 years ago
An atom of aluminum has 13 protons and 14 neutrons. what is the atomic mass
PtichkaEL [24]
A=P +N
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5 0
3 years ago
At equilibrium the reactant and product concentrations are constant because a change in one direction is balanced by a change in
Anna [14]

Answer: a. The concentrations of the reactants and products have reached constant values

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.  For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equal to rate of the backward reaction.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time and are constant.

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For a equilibrium reaction,

A\rightleftharpoons B

K_{eq}=\frac{[B]}{[A]}

Thus the correct answer is the concentrations of the reactants and products have reached constant values.

5 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(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_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

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_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

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 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
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