Answer:
ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)
Explanation:
Our strategy in this question is to use the trend in entropies :
S (solids) less than S (liquids) less than S (gases)
Also we have to look for the molar quanties involved of each state and their change to answer the question:
A. N2(g) + 3 H2(g) → 2 NH3(g)
Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.
B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)
The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.
C. CH3OH(l) → CH3OH(s)
A liquid has a higher entropy than a solid so ΔS is negative
D. False see A,B,C
E. The change in moles of gases is 4 - 2= 2, therefore ΔS is greater than O.
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =
Now put all the given values in the formula of molality, we get
According to the dilution law,
where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:
Therefore, the concentration of KOH for the final solution is 0.275 M
<h3>Answer:</h3>
6 moles of H₂O / 9 moles of O₂
<h3>Explanation:</h3>
The Balance Chemical Equation is as follow,
2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O
According to this balance equation 2 moles of C₃H₆ (Propene) when combusted with 9 moles of O₂ (Oxygen) produces 6 moles of CO₂ (Carbon Dioxide) and 6 moles of H₂O (Water Vapors).
Hence, Given options are;
1 mole of C₃H₆ / 2 moles of CO₂: This is incorrect because 1 mole of C₃H₆ will produce 2 moles of CO₂.
6 moles of H₂O / 9 moles of O₂: This is correct as 6 moles of H₂O is produced by reacting 9 moles of O₂.
2 moles of C₃H₆ / 6 moles of O₂: This ratio is incorrect because for complete oxidation 2 moles of C₃H₆ requires 9 moles of O₂.
3 moles of H₂O / 2 moles of CO₂: This is also incorrect because the production of 3 moles of H₂O will be accompanied by the production of 3 moles of CO₂.