Answer:
1) increase
2) increase
Explanation:
Entropy is the degree of disorderliness or randomness of a system. It is the measure of the unavailable energy in a system.
Entropy increases with increase in the number of particles. If the number of particles in a system increases from left to right, the entropy of the system increases accordingly.
In reaction 1, the number of particles from left to right increased from two to three hence the entropy was increased.
In reaction 2, the number of particles from left to right increased from three to five hence the entropy was increased.
<span>When an ice cube is placed on a kitchen counter, heat will flow from the ice cube to the counter, causing the molecules in the counter to move more slowly. The molecules of the counter move more slowly because the heat transferred to them from the ice has reduced their kinetic energy.</span>
Outcome (dependent) variable-drinking energy drinks
Test (independent) variable- more aggressive
Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
We know that:
Molar Mass H2O: 18 g/mol
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>
<span>Using formula:
V= [mole fraction x molar mass] / density </span>
<span>mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 </span>
<span>morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 </span>
<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>
Yotal volume = 30 mL; therefore,
<span>0.0442 = (volume eugenol/30) </span>
<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
= 1.44/1.05
= about 1.37 mL </span>
