Answer:
0.66atm
Explanation:
Based on the reaction:
NH₄HS(s) ↔ NH₃(g) + H₂S(g)
The equilibrium constant, Kp, is defined as:
Kp = 0.11 = 
As moles of gas produced for NH₃(g) and H₂S(g) are the same, it is possible to write:
That means pressure of NH₃(g) is 0.33atm and H₂S(g) is, also, 0.33atm. Thus, total pressure is:
0.33atm×2 = <em>0.66atm</em>
Answer: 3
Explanation:
An oxide-reduction reaction or, simply, redox reaction, is a <u>chemical reaction in which one or more electrons are transferred between the reactants</u>, causing a change in their oxidation states, which is the hypothetical electric charge that the atom would have if all its links with different elements were 100% ionic.
For there to be a reduction-oxidation reaction, in the system there must be an element that yields electrons and another that accepts them:
-
The oxidizing agent picks up electrons and remains with a state of oxidation inferior to that which it had, that is, it is reduced.
- The reducing agent supplies electrons from its chemical structure to the medium, increasing its oxidation state, ie, being oxidized.
To balance a redox equation you must <u>identify the elements that are oxidized and reduced and the amount of electrons that they release or capture, respectively.
</u>
In the reaction that arises in the question the silver (Ag) is reduced <u>because it decreases its oxidation state from +1 to 0</u> and the aluminum (Al) is oxidized because <u>its oxidation state increases from 0 to +3</u>, releasing 3 electrons (e⁻). Then we can raise two half-reactions:
Ag⁺ + e⁻ → Ag⁰
Al⁰ → Al⁺³ + 3e⁻
In order to obtain the balanced equation, we must multiply the first half-reaction by 3 so that, when both half-reactions are added, the electrons are canceled. In this way:
(Ag⁺ + e⁻ → Ag⁰ ) x3
Al⁰ → Al⁺³ + 3e⁻ +
-------------------------------------
3Ag⁺ + Al⁰ → 3Ag⁰ + Al⁺³
So, the coefficient of silver in the final balanced equation is 3.
Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:
According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:
hydrocarbon is ethene which is used to test for saturation and it undergoes addition reaction
