A mole is a unit to describe an amount of something/specific substance. Typically standard for measuring a large quantity of small entities like atoms, molecules, or other particles.
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
Answer:

Explanation:
Hello!
In this case, since the decomposition of sodium hydrogen carbonate is:

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

Best regards!
So I’m not 100% sure what you’re asking but I’m going to give it a go. The elimination reaction is a term used in organic chemistry that describes a type of reactions. The name kinda tells you what’s going to happen. Something is going to be removed/eliminated from initial reactant/substrate and as a result, an alkene (double bond containing compound) will form.
In elimination reactions a hydrogen atom is first removed (as a H+) from the beta carbon. As a result, the left behind electrons create a pi bond between the beta carbon and the neighboring alpha carbon. This promotes the electronegative atom, on the alpha carbon, to leaves the substrate taking both electrons from the shared sigma bond with the alpha carbon.