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
The larger and heavier the box, the more force you must use when pushing.
Answer:
i think the answer is letter C. From 35°c to 45°c
Explanation:
sorry if it is wrong
Answer:
40%
Explanation:
We'll begin by obtaining the molar mass of MgO. This is illustrated below:
Molar Mass of MgO = 24 + 16 = 40g/mol
Observing the formula MgO, we have 1 atom of O in it.
The percentage composition by mass of oxygen in MgO is given by:
Mass of O/Molar Mass of MgO x 100
= 16/40 x 100 = 40%
The answer to your question is D. Electrolysis.
