Answer:
The heat of formation of SO2 is -296.1 kJ
Explanation:
<u>Step 1:</u> Data given
2S (s)+3O2 (g)→2SO3 (g) ΔH = -790.4 kJ
2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ
<u>Step 2</u>: Calculate the heat of formation of SO2
2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ
S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ
2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ
SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ
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S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ
SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ
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S (s)+O2 (g)→SO2 (g)
ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ
ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ
The heat of formation of SO2 is -296.1 kJ
<u>Answer:</u> The temperature of the solution in Kelvins is 422.356 K
<u>Explanation:</u>
Temperature is defined as the measure of coldness or hotness of a body. It also determines the average kinetic energy of the particles in a body.
This term is expressed in degree Celsius, degree Fahrenheit and Kelvins. All these units are interchangeable.
The S.I unit of temperature is Kelvins.
We are given:
Temperature of a solution = 
Conversion used to convert degree Celsius and Kelvins is:
![T(K)=[273.15+T(^oC)]](https://tex.z-dn.net/?f=T%28K%29%3D%5B273.15%2BT%28%5EoC%29%5D)
Hence, the temperature of the solution in Kelvins is 422.356 K
All matter is made up of substances called elements, which have specific chemical and physical properties and cannot be broken down into other substances through ordinary chemical reactions. Gold, for instance, is an element, and so is carbon. There are 118 elements, but only 92 occur naturally.
<h3>
<u>P</u><u>LEASE </u><u>MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>
Answer:
Bueno,
Explanation:
Tres pesnamientos negativos sobre el C0Vid-19 son,
- Encerrados en la casa
- Tener que ponerse mascara
- Todos los lugares divertidos cerrados
Convertidos en positivos:
- Pasamos mas tiempo de calidad en familia.
- Nos protejemos unos a otros.
- Descansan un poco todas las personas que se han pasado toda la vida trabajando.
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
