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Brrunno [24]
3 years ago
8

On the basis of the solubility curves shown above, the greatest percentage of which compound can be recovered by cooling a satur

ated solution of that compound from 90°C to 30°C? A) KCl B) NaNO3 C) NaCl D) KNO3
Chemistry
2 answers:
lys-0071 [83]3 years ago
6 0
The answer is D) KNO3
Nookie1986 [14]3 years ago
6 0
The answer is D) <span> <span> KNO3 </span> </span>
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5. Durante un estudio de la velocidad de la reacción A2(g) + 3B2(g)  2 AB3(g), se observa que en un recipiente cerrado que cont
weqwewe [10]

Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

Hope this Helps!!!

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Answer:

We'll have 82 moles ZnO and 41 moles S

Explanation:

Step 1: data given

Number of moles Zinc (Zn) = 82 moles

Number of moles sulfur oxide (SO2) = 42 moles

Step 2: The balanced equation

2Zn + SO2 → 2ZnO + S

Step 3: Calculate the limiting reactant

For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur

Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles

There will remain 42-41 = 1 mol SO2

Step 4: Calculate moles of products

For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur

For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)

For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur

We'll have 82 moles ZnO and 41 moles S

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