Answer:
2
Explanation:
#1 is incorrect as there is no upward force that would balance the 5 N downward force
#3 is incorrect 2 is less than 5 = net force downward of 3 N
#4 is incorrect 5 is greater than three = 2 N net force going to the left and 2N going up
Answer:
Mass of barium sulfate = 8.17 g
Explanation:
Given data:
Mass of sodium sulfate = 4.98 g
Mass of barium sulfate produced = ?
Solution:
Na₂SO₄ + Ba(NO₃)₂ → BaSO₄ + 2NaNO₃
Moles of sodium sulfate:
Number of moles = mass/molar mass
Number of moles =4.98 g / 142.04 g/mol
Number of moles = 0.035 mol
Now we will compare the moles pf sodium sulfate and with barium sulfate.
Na₂SO₄ : BaSO₄
1 : 1
0.035 : 0.035
Mass of barium sulfate:
Mass = number of moles × molar mass
Mass = 0.035 mol ×233.4 g/mol
Mass = 8.17 g
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|.
