The answer would be C as in cat and D as in dog your welcome i hope this help
Less dense than the liquid
Answer:
B: Adding water, then adding solute
Explanation:
This is because, say you have a solution with a certain concentration.
If you add more water, it will become more diluted (less concentrated)
If you add more solute, it will become more concentrated.
Therefore if you add water and solute, it could cancel out, and the concentration would remain the same.
Hope this helps! Let me know if you have any questions/ would like anything further explained :)
Answer:
Entropy increases
Explanation:
Entropy (S) is a measure of the degree of disorder. For a given substance - say water - across phases the following is true ...
S(ice) < S(water) << S(steam)
For a chemical process, entropy changes can be related to increasing or decreasing molar volumes of gas from reactant side of equation to product side of equation. That is ...
if molar volumes of gas increase, then entropy increases, and
if molar volumes of gas decrease, then entropy decreases.
For the reaction 2KClO₃(s) => 2KCl(s) + 3O₂(g)
molar volumes of gas => 0Vm* 0Vm 3Vm
*molar volumes (Vm) apply only to gas phase substances. Solids and liquids do not have molar volume.
Since the reaction produces 3 molar volumes of O₂(g) product vs 0 molar volumes of reactant, then the reaction is showing an increase in molar volumes of gas phase substances and its entropy is therefore increasing.
Answer:
B) -1551 kJ
Explanation:
There are two heat flows in this question.
Heat released by engine + heat absorbed by water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m = 2.51 kg
C = 3.41 J°C⁻¹g⁻¹
T_i = 23.8 °C
T_f =205 °C
Calculations:
(a) ΔT
ΔT = 205 °C - 23.8 °C = 181.2 °C
(b) q₂
q₂ = mCΔT = 2510 g × 3.41 J·°C⁻¹g⁻¹ × 181.2 °C = 1.551× 10⁶ J = 1551 kJ
(c) q₁
q₁ + 1551 kJ = 0
q₁ = -1551 kJ
