Answer:
3,116J/K
Explanation:
This question asks to calculate the entropy change of the surroundings.
To do this, we need the standard enthalpy of formation ΔfH° of the reacting species and products first:
We should observe that standard enthalpy if formation of O2 is zero. We proceed with the rest of the species.
H2CO = -109.5KJ/mol
CO2 = -393.5KJ/mol
H2O = -285.8KJ/mol
Now, we calculate the standard change of enthalpy of the reaction as:
ΔHrxn = ΔHproduct - ΔHreactant = (-285.8 - 393.5) +(109.5) = -569.8 KJ/mol
The relationship between the entropy and the standard formation enthalpy is given as
The relationship is:
ΔSosurroundings = - ( ΔHof/ T)
We convert the standard enthalpy of formation to joules first = -569.8 * 10^3 Joules
Using the formula above at a temperature of 298k, the entropy change would be:
-(-569.8 * 10^3)/298 = 1912J/K
Now, we know that 1.63 moles of H2CO reacted. We also need to know the coefficient of the H2CO in the reaction which is 1.
We thus have:
1.63 mol H2CO(g) * (1912J/K * 1 mol H2CO) = 3116J/K
D is true because NO2 is a catalyst, and a catalyst is a compound which makes a reaction occur faster (not necessarily more effectively, creating more product, but it decreases the amount of time it takes).
The energy of an electron can be change only by set amounts.
The electrons in an energy level of an atom can only move to a higher or lower energy level only if it absorbs or emits a certain amount of energy. If the electron does not have enough energy to move to the next energy level then it will never move.
Explanation:
There are 6.022x1023 molecules in 1 mole of SO3 (Avogadro's number) so in 0.25 moles, there are (0.25)(6.022x1023) molecules in the 0.25 moles of SO3. or 1.506x1023 molecules per 0.25 moles of SO3.
There is one atom of sulfur in each molecule of SO3. So there are 1.506x1023 atoms of sulfur in 0.25 moles of SO3.
