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!
Always remember that pH + pOH = 14
Here, you have a pOH of 11.24, so you replace it in the equation, and u get:
pH + 11.24 = 14
Then, You move 11.24 to the other part. and moving from a part to another change the sign of the equation. And you get:
pH = 14 - 11.24 = 2.76
So, the pH of a solution that has a pOH of 11.24 is pH = 2.76
Hope this Helps :)
Answer:
The core of Jupiter and Saturn is made up of rock,metal and hydrogen compounds, while the core of Uranus and Neptune is made up of rock,metals,water, methane, and ammonia.
Explanation:
The Jovian planets include, Jupiter, Saturn, Uranus, and Neptune. These planets when compared to terrestrial planets are small, with dense cores and surrounded by layers of gas.
The solubility of a sample will DECREASE when the size of the sample increases.
The bigger a substance is, the more will be the particles that make up this substance and the greater the amount of solvent that will be needed to dissolve the substance. Surface area of the substance is also important, a small surface area will impede solubility. Thus, when the size of a sample increases, the solubility decreases.
