All categories

2 years ago

Determine the pH, pOH, [H+], and [OH−] of a solution in which 0.300 g of aluminum hydroxide is dissolved in 184 mL of solution.

Chemistry

1 answer:

lys-0071 [83]2 years ago

3 0

Answer:

[OH⁻] = 0.0627M

pOH = 1.20

pH = 12.8

[H⁺] = 1.59x10⁻¹³M

Explanation:

To solve this question we must, as first, find the molarity of Al(OH)₃ in the solution -Molar mass Al(OH)₃: 78.00g/mol-:

0.300g * (1mol/ 78.00g) = 3.846x10⁻³ moles

In 184mL = 0.184L:

3.846x10⁻³ moles / 0.184L = 0.0209M Al(OH)₃. Three times this molarity = [OH⁻]:

[OH⁻] = 0.0209M * 3

pOH = -log [OH⁻] =

pH = 14 - pOH

And [H⁺] = 10^-pH

You might be interested in

I need help on this plsss

Firlakuza [10]

Answer:

An engine that creates ignites fuel with highly compressed air.

6 0

3 years ago

Read 2 more answers

You could also graph decay time (in seconds or minutes) vs. number of radioactive nuclei. How would this graph differ from the g

Tatiana [17]

Only one of the listed choices are correct here:
<span><em>The x-axis would change title and values.</em>
</span>

6 0

3 years ago

Read 2 more answers

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0

3 years ago

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!!!

3 0

2 years ago

How much heat is released when you condense 93.9 g of water vapor? <br><br> Show work please ❤️

sammy [17]

<u>Answer:</u>

211.9 J

<u>Explanation:</u>

The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.

The amount of heat released can be represented by the formula:

$Q=mL_e$

where $Q$ = heat energy, $m$ = mass of water and $L$ = latent heat of evaporation.

The latent heat of evaporation for water is $L=2257 kJ/kg$ and the mass of the water is $m=93.9 g=0.0939 kg$ .

The amount of heat released in this process is:

$Q=mL_e = (0.0939kg)(2257 kJ/kg)=$ 211.9 J

7 0

3 years ago