Answer:

Explanation:
Hello,
In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

Best regards.
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
<span>First, write the net ionic equation for the unbalanced reaction. If you are given a word equation to balance, you'll need to be able to identify strong electrolytes, weak electrolytes and insoluble compounds. Strong electrolytes completely dissociate into their ions in water. Examples of strong electrolytes are strong acids, strong bases, and soluble salts. Weak electrolytes yield very few ions in solution, so they are represented by their molecular formula (not written as ions). Water, weak acids, and weak bases are examples of weak electrolytes. The pH of a solution can cause them to dissociate, but in those situations, you'll be presented an ionic equation, not a word problem. Insoluble compounds do not dissociate into ions, so they are represented by the molecular formula. A table is provided to help you determine whether or not a chemical is soluble, but it's a good idea to memorize the solubility rules.
</span><span><span>arate the net ionic equation into the two half-reactions. This means identifying and separating the reaction into an oxidation half-reaction and a reduction half-reaction. </span><span>For one of the half-reactions, balance the atoms except for O and H. You want the same number of atoms of each element on each side of the equation. </span><span>Repeat this with the other half-reaction. </span><span>Add H2O to balance the O atoms. Add H+ to balance the H atoms. The atoms (mass) should balance out now. </span><span>Now balance charge. Add e- (electrons) to one side of each half-reaction to balance charge. You may need to multiply the electrons the the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation. </span><span>Now, add the two half-reactions together. Inspect the final equation to make sure it is balanced. Electrons on both sides of the ionic equation must cancel out. </span><span>Double-check your work! Make sure there are equal numbers of each type of atom on both sides of the equation. Make sure the overall charge is the same on both sides of the ionic equation. </span><span>If the reaction takes place in a basic solution, add an equal number of OH- as you have H+ ions. Do this for both sides of the equation and combine H+ and OH- ions to form H2O. </span><span>Be sure to indicate the state of each species. Indicate solid with (s), liquid for (l), gas with (g), and aqueous solution with (aq). </span><span>Remember, a balanced net ionic equation only describes chemical species that participate in the reaction. Drop additional substances from the equation.ExampleThe net ionic equation for the reaction you get mixing 1 M HCl and 1 M NaOH is:H+(aq) + OH-(aq) → H2O(l)Even though sodium and chlorine exist in the reaction, the Cl- and Na+ ions are not written in the net ionic equation because they don't participate in the reaction.</span></span>