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sineoko [7]
3 years ago
11

Why is gold usually found in its pure form

Chemistry
1 answer:
attashe74 [19]3 years ago
5 0

gold is usually found in pure form because it is not reacting with other chemicals naturally.
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Explanation:

Antibiotics are chemical agents that are used to treat bacterial infections. ... Vaccines are used to prevent infection, particularly viral infections.

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El nivel simbólico representa tanto a lo macroscópico como a lo nanoscópico.
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verdadero

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Water is a polar solvent. Which of the following will not dissolve in water? (3 points) 3Points A sugar B table salt C vinegar D
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The answer is D Olive Oil

Due to water being a polar substance, and olive oil is a non-polar substance. They will not mix, the olive oil will sink to the bottom of the water and not dissolve.

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How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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What are the three properties of gas
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