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Brrunno [24]
3 years ago
12

An expression is show below. - 1 2/3 - x

Chemistry
1 answer:
natka813 [3]3 years ago
4 0

Answer:

What’s the answer ‍♀️

Explanation:

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Why does nobody fking answer my questions.
raketka [301]

Answer:

by wearing of rocks

Explanation:

earths gravity your welcome

6 0
2 years ago
Read 2 more answers
Why burning of fuel is a chemical change give 5 reasons to support your answer
frozen [14]

Answer:

Explanation:

1. New substances such as carbondioxide and water is formed.

2. There is evolution of gas bubbles. Gases are released.

3. There is either the absorption of energy or release of energy in form of light and energy.

4. The reaction is irreversible i.e it cannot be reversed, it is permanent once the reaction take place.

5. There is a change in both odor and smell.

A chemical change is a change where new substances are formed due to changes in the properties.

6 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
2 years ago
What is element Na in a periodic table
Bess [88]

Answer:

Sodium.  

Explanation:

Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table. Its only stable isotope is ²³Na.

7 0
3 years ago
Read 2 more answers
(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b)
Gala2k [10]

Answer:

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻  with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺

HOAc delivers more H⁺ than HClO and is more acidic.

Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

4 0
3 years ago
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