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
Beryllium (Be) and Calcium (Ca)
The Nuclear Model-1909. Ernest Rutherford's great contribution to atomic theory was to show what happens to an element during radioactive decay. This enabled him to construct the first nuclear model of the atom, a cornerstone of present-day physics The Nuclear Model-1909.
Answer:
23 kgs = 809.6 ounces
Explanation:
given:
1 kg = 2.2 ibs
therefore, 23 kgs = 2.2*23 Ibs = 50.6 Ibs
since, 1 Ib = 16 ounces
therefore, 23 kgs = 50.6*16 ounces = 809.6 ounces
Answer:
The core elements are hydrogen and helium.
Explanation:
