Answer: 6.02214179×1023 atoms
Explanation: I think so
The first thing to do is to convert the mass of sodium
sulfate to moles. Since the given is in milligrams convert it to grams which
would be 0.02824 grams. The number of moles can be calculated by dividing the
converted weight with the given molar mass which is 139.04 g/mol. The number of
mols in the solution is 2.03x10^-4 mols. Then divide the number of moles with
0.02355 L. The molarity of the solution is 0.00862 M.
Electrical energy changes to heat energy
Answer:
A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
Explanation:
The spontaneity of a reaction depends on the Gibbs free energy(ΔG).
- If ΔG < 0, the reaction is spontaneous.
- If ΔG > 0, the reaction is nonspontaneous.
ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T.ΔS
where,
T is the absolute temperature (always positive)
Regarding the exchange of heat:
- If ΔH < 0, the reaction is exothermic.
- If ΔH > 0, the reaction is endothermic.
<em>Which statement is true? </em>
<em>A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic. </em>TRUE. If ΔS < 0, the term -T.ΔS > 0. ΔG can be negative only if ΔH is negative.
<em>A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.</em> FALSE. If ΔS > 0, the term -T.ΔS < 0. ΔG can be negative if ΔH is negative.
<em>A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.</em> FALSE. If ΔS < 0, the term -T.ΔS > 0. ΔG cannot be negative if ΔH is positive.
<em>A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.</em> FALSE. If ΔS > 0, the term -T.ΔS < 0. ΔG can be negative even if ΔH is positive, as long as |T.ΔS| > |ΔH|.
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
