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ICE Princess25 [194]

3 years ago

9

Consider the following chemical reaction of bromothymol blue indicator. It appears yellow in undissociated form and blue in its

dissociated aqueous solution. HC2H3O2(aq) Double headed arrow. H (aq) C2H3O2–(aq) yellow blue What will be the color of the solution if a large amount of H2CO3 is added? The solution will remain yellow. The solution will turn blue. The solution will turn pink. The solution will turn green.

1 answer:

TiliK225 [7]3 years ago

6 0

The addition of hydrogen carbonate to bromothymol blue turns the solution blue. Thus, option B is correct.

The balanced equation for the dissociation of bromothymol blue is:

$\rm HC_2H_3O_2\;\leftrightharpoons H^+\;+\;C_2H_3O_2^-$

The color of dissociated form is yellow and undissociated form is blue.

<h3>What is the final color of solution?</h3>

The addition of hydrogen carbonate results in the dissociated ions as:

$\rm H_2CO_3\;\rightarrow\;2\;H^+\;+\;CO_3^-$

The dissociation results in the increased hydrogen ion concentration. The undissociated form in the reaction mixture increases.

Thus, the color of the solution will turn blue. Hence, option B is correct.

Learn more about bromothymol blue, here:

brainly.com/question/24319054

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Explanation:

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The molal freezing point constant for copper is 23 °C/m. If pure copper melts at 1083°C, what will be the melting point of a bra

nikitadnepr [17]

<u>Answer:</u> The freezing point of brass is 1028.57°C

<u>Explanation:</u>

We are given:

13.4 mass percent of zinc in brass

This means that 13.4 grams of zinc is present in 100 g of brass

Mass of copper = [100 - 13.4] g = 86.6 g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure copper = 1083°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 23°C/m

$m_{solute}$ = Given mass of solute (zinc) = 13.4 g

$M_{solute}$ = Molar mass of solute (zinc) = 65.38 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 86.6 g

Putting values in above equation, we get:

$1083-\text{Freezing point of solution}=1\times 23^oC/m\times \frac{13.4\times 1000}{65.38g/mol\times 86.6}\\\\\text{Freezing point of solution}=1028.57^oC$

Hence, the freezing point of brass is 1028.57°C

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3 years ago

Which of the following is a homogeneous mixture? a. Na(s) b. H2O(l) c. Cl2(g) d. NaCl(aq)

iVinArrow [24]

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3 years ago

At 36°C, what is the osmotic pressure of a 0.82% NaCl by weight aqueous solution? Assume the density of the solution is 1.0 g/mL

Setler79 [48]

Answer: e. 3.5 atm

Explanation:

$\pi =CRT$

$\pi$ = osmotic pressure = ?

C= concentration in Molarity

R= solution constant = 0.0821 Latm/Kmol

T= temperature = $36^0C=(36+273)K=309K$

For the given solution: 0.82 grams of $NaCl$ is dissolved in 100 g of solution.

${\text {volume of solution}}=\frac{\text {mass of solution}}{\text {Density of solution}}=\frac{100g}{1.0g/ml}=100ml$

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

Putting in the values we get:

$C_{NaCl}=\frac{0.82\times 1000}{58.5\times 100}=0.14M$

$\pi=0.14mol/L\times 0.0821Latm/Kmol\times 309K$

$\pi=3.5atm$

The osmotic pressure of a 0.82% NaCl by weight aqueous solution is 3.5 atm

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