Part 1 : Answer is only B substance is soluble in water.
In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.
Part 2 : Substance A is not soluble in water and substance B is soluble in water.
According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.
But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.
Answer:
The concentration of acetic acid in vinegar of that trial would be <u><em>greater than</em></u> the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
3.25 kg in g = 3.25 * 1000 = 3250 g
Molar mass C₂H₆O₂ = 62.0 g/mol
Mass solvent = 7.75 kg
Number of moles:
n = mass solute / molar mass
n = 3250 / 62.0
n = 52.419 moles
Molality = moles of solute / kilograms of solvent
M = 52.419 / 7.75
M = 6.7637 mol/kg
hope this helps!
Answer:
B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]
Explanation:
For the reaction A + B → C
The formula for rate of reaction is:
Δ[C]/Δt = k [A] [B]
As you have [A], [B] and Δ[C]/Δt information you can multiply [A] times [B] and take this value as X and Δ[C]/Δt as Y. The slope of this lineal regression will be k.
Thus, you must obtain:
y = 3,60x10⁻² X
Thus, rate of reaction is:
B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]
I hope it helps!
