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Vladimir79 [104]
3 years ago
10

Which substances have AH = 0 kJ/mol

Chemistry
1 answer:
aniked [119]3 years ago
7 0

Answer:

Oxygen, bromine, iron, helium

Explanation:

\Delta H^o_f is defined as the standard enthalpy of formation. By definition, the standard enthalpy of formation is equal to 0 kJ/mol for the substances in their standard states, that is, at room temperature and 1 atm pressure.

Simply speaking, looking at the substances given, we need to understand whether their states agree with what we expect to see at standard conditions (e. g., sodium is a metal, fluorine is a gas, bromine is a liquid at standard conditions). Those are substances consisting of just one type of atoms.

  • Firstly, oxygen is a gas at standard conditions and it is diatomic, so its \Delta H^o_f=0 kJ/mol.
  • Although nitrogen is a gas at standard conditions, it is diatomic, so \Delta H^o_f\neq 0 kJ/mol.
  • Water is a liquid at standard conditions, however, it consists of two types of atoms, hydrogen and oxygen, so \Delta H^o_f \neq 0 kJ/mol.
  • Bromine is a liquid at standard conditions, so \Delta H^o_f=0 kJ/mol.
  • Iron is a solid at standard conditions, it's a metal, so \Delta H^o_f=0 kJ/mol.
  • Helium is a gas at standard conditions, it belongs to noble gases, so \Delta H^o_f=0 kJ/mol.
  • Sulfur is a solid at room conditions, however, the conformation it has is S_8 and not S_6, so \Delta H^o_f\neq 0 kJ/mol.
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An ion of oxygen- 16 contains 8 protons and has a 2- charge. How many electrons does it have?
Verdich [7]

Answer:

i would say 10, so the anser is A.

Explanation:

because there are the same number of protons and electrons, therefore for a regular O, you are supposed to have only 8 protons, but it is charged, thus, whatever the charge is will be taken into consideration into how much the proton and electron doe it have. Thus, for this case, it has 10, because the charge is negative and you have 8 electron plus 2 = 10.

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Coal can be used to generate hydrogen gas (a potential fuel) by the following endothermic reaction:C(s)+H2O(g) ------------->
tia_tia [17]

Answer:

A. No effect

B. Results in the formation of additional hydrogen gas

C. Results in the formation of additional hydrogen gas

D. Results in the formation of additional hydrogen gas

E. No effect

F. No effect

Explanation:

The equilibrium in this question is

C(s) + H₂O (g) ⇄ CO(g) + H₂ (g)

and

Kp = pCO x pH₂/ pH₂O

where pCO, pH₂O and pH₂O are the partial pressures of CO, H₂ and H₂O.

We call the equilibrium constant Kp since only gases intervene in the expression for the constant.

A. adding more C to the reaction mixure

Adding more carbon which is a solid does not alter the  pressure equilibrium constant, therefore, it has no effect on the equilibrium and consequently no effect on the quantity of hydrogen gas.

B. adding more H₂O to the reaction mixture

We can answer this part by using  Le Chatelier's principle which states that a system at equilibrium will respond to a stress in such a way as to minimize the stress, hence  restoring equilbrium.

One of the three possible stresses is an increase of reactant as in this case. The system will react by decreasing some of the added water. Thus the equilbrium shifts to the product side which will result in the formation of more hydrogen gas.

The difference of this part with respect to part A is that indeed the water gas is included in the equilibrium constant expression.

C. raising the temperature

This is another stress we can subject an equilibrium.

We are told the reaction is endothermic which means in going from left to right it consumes heat. Thus the equilibrium will shift to the product side by consuming some of the added heat favoring the production of more hydrogen gas.

D. increasing the volume of the reaction mixture

This the last of the stresses .

Increasing the volume of the reaction effectively decreases the pressure ( volume is inversely proportional to pressure ) so the equilibrium will shift to the side that has more pressure which is the product side: we have two moles of gases  products  vs. 1 mol gas in the reactant side.

Therefore, the equilibrium will shift to the right increasing the quantity of H₂.

E. adding a catalyst to the reaction mixture

The addition of a catalyst does not have an effect on the equilibrium constant. The catalyst will speed both the forward and reverse reaction decreasin the time to attain equilibrium.

So there is no effect on the quantity of H₂.

F. Adding an inert gas to reaction mixture

Assuming the volume of the reaction mixture remains constant, and we are not told such change in volume occurred, the addition of an inert gas does not have an effect in our equilibrium. The inert gasdoes not participate  in the calculation for Kp.

The situation will be different if the volume of the reaction is allowed to increase, but again this is not stated in the question.

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What is meant by the biodegradability of soap molecules
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How many moles of CO2 are produced from 15.6 g of NaHCO3?
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Answer:

0.185moles

Explanation:

1mole of bicarbonate produces 1mole of CO2

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How do you the density of a matchbox car which has a mass of 10 grams and the water level in the cylinder rises from 41 ml to 46
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Answer:

I don't know how to label it, but the answer is 5.

Explanation:

The answer is 5 because the rock made the water move up 5 mL.

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