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12345 [234]
2 years ago
13

Score! You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some o

f the first unlabeled solution and some indicator. The color changes to yellow. You then add some solution from the other unlabeled flask into this beaker and see the color change to blue. What are the identities of each unlabeled solution
Chemistry
1 answer:
mamaluj [8]2 years ago
6 0

Answer:

You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some of the first unlabeled solution and some indicator.

The color changes to yellow.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

What are the identities of each unlabeled solution?

Explanation:

Bromothymol blue is a dye and it is used as an indicator.

It is used as a pH indicator.

In acids, it becomes yellow n in color.

In bases, it turns blue.

You take one of the empty beakers and add some of the first unlabeled solution and some indicator. The color changes to yellow.

That means the unlabeled solution is an acid.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

It is a basic solution.

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A 57.0 mL sample of a 0.120 M potassium sulfate solution is mixed with 35.5 mL of a 0.118 M lead(II) acetate solution and the fo
Katarina [22]

Answer:

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

Moles=0.120 M \times {57.0\times 10^{-3}}\ moles

Moles of potassium sulfate  = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

Moles=0.118 \times {35.5\times 10^{-3}}\ moles

Moles of lead(II) acetate  = 0.004189 moles

According to the given reaction:

K_2SO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow 2KC_2H_3O_2_{(s)}+PbSO_4_{(aq)}

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %</u>

3 0
3 years ago
Which event is most likely an example of a chemical reaction
Inessa [10]

Explanation:

One example of a chemical reaction is the rusting of a steel garbage can. That rusting happens because the iron (Fe) in the metal combines with oxygen (O2) in the atmosphere. Chemical bonds are created and destroyed to finally make iron oxide (Fe2O3).

7 0
3 years ago
Connect and Apply scientific knowledge to explain the following situations:
mixer [17]

Answer:

b.Rearrange the two equations to make each variable the focus (you will end up with three variations for each equation).

7 0
2 years ago
What is the ph of a 0.45 m solution of aniline (c6h5nh2)? (pkb  9.40)?
mote1985 [20]

Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+ H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).

pKb(C₆H₅NH₂) = 9.40.

Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.

c₀(C₆H₅NH₂) = 0.45 M.

c(C₆H₅NH₃⁺) = c(OH⁻) = x.

c(C₆H₅NH₂) = 0.45 M - x.

Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).

4·10⁻¹⁰ = x² /  (0.45 M - x). 

Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.

pOH = -log(0.0000134 M.) = 4.87.

pH = 14 - 4.87 = 9.13.

7 0
3 years ago
What is the hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4.
Maksim231197 [3]

The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7.

pH= -log[H+] - (i)

10^-3=H2So4

H+= 2×10-3

here ,

h2so4 ——— 2[H+] + so4^2-

thus [H+]= 2*10^(-3) because hydrogen ion has two moles

pH= -log[H+]

pH= -log(2×10^-3)

pH= 3-log2

pH= 3-log2pH= 2.7

The pH is 2.7

<h3>What is pH?</h3>

PH is the degree of alkalinity and acidicity in a solution.

Therefore, The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7

Learn more about pH from the link below.

https://brainly.in/question/9937410

4 0
2 years ago
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