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:
<em>The electrons in an atom can only occupy certain allowed energy levels to a lower one</em>, the excess energy is emitted as a photon of light, with its wavelength dependent on the change in electron energy. This is why an atom can only emit specific wavelengths of light and not every possible wavelength.
The final step in a typical titration, that is here an acid base one would be to finally find the concentration of your unknown substance whether that be the acid or the base. The other steps are used before this to come to the correct calculation and conclusion.
Answer:
Final temperature =
Explanation:
Given that,
Heat added, Q = 250 J
Mass, m = 30 g
Initial temperature, T₁ = 22°C
The Specific heat of Cu= 0.387 J/g °C
We know that, heat added due to the change in temperature is given by :
Put all the values,
So, the final temperature is equal to .
The mass number of an isotope can be expressed <span>by simply writing the name of the element or symbol followed by a hyphen and the mass number.
Example:
Carbon-13, Carbon-14
Oxygen-17
Uranium-235
</span>