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3 years ago

11

What color change is exhibited by phenolphthalein during a titration of aqueous acetic acid with aqueous sodium hydroxide ? a. c

olorless to pink b. pink to colorless c. no color change d. yellow to blue e blue to yellow

1 answer:

Agata [3.3K]3 years ago

4 0

Phenolphthalein during a titration

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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

4 years ago

Which of the following is the primary medium for beach erosion?

Rus_ich [418]

Answer:

Water is the primary erosion

hope it helps :)

3 0

4 years ago

Would any Chemistry braniacs be kind enough to help explain how to work these out? Any help is appreciated, thanks

o-na [289]

Answer:

NO.3) Mass of Al2O3 formed = 229.5g

7 0

3 years ago

A student reacts 20.0 grams of sodium hydroxide with excess hydrochloric acid, and obtains a

Tcecarenko [31]

Answer:

Percent yield of NaCl = 87.75 %

Explanation:

Data given

mass of sodium = 20.0 g

actual yield of Sodium chloride = 25 g

percent yield of sodium chloride = ?

Reaction Given:

NaOH + HCI --> NaCl + H₂O

Solution:

First we have to find theoretical yield.

So, Look at the reaction

NaOH + HCI ---------> NaCl + H₂O

1 mol 1 mol

As 1 mole of NaOH give 1 mole of NaCl

Convert moles to mass

molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

molar mass of NaCl = 23 +35.5 = 58.5 g/mol

Now

NaOH + HCI ---------> NaCl + H₂O

1 mol (40 g/mol) 1 mol (58.5 g/mol)

40 g 58.5 g

So 40 g of NaOH gives 58.5 g of NaCl so how many grams of Sodium chloride will be produced by 20 g of NaOH.

Apply Unity Formula

40 g of NaOH ≅ 58.5 g of NaCl

20 g of NaOH ≅ X g of NaCl

Do cross multiply

g of NaCl = 58.5 g x 20 g / 40 g

g of NaCl = 29.25 g

So the Theoretical yield of NaCl = 29.25 g

Now Find the percent yield of NaCl

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 25 / 29.2 x 100 %

percent yield = 87.75 %

percent yield of NaCl = 87.75 %

7 0

4 years ago

An unknown metal displaces cadmium (Cd) from solution but does not displace chromium (Cr). Using the activity series, determine

sdas [7]

Answer: The correct answer is Iron (Fe).

Explanation:

When a metal displaces another metal from a solution, it is considered as a reactive metal and the reactivity of metal is judged by the reactivity series of the metal.

Whenever one metal displaces another metal from its salt solution, it is considered as a Single-displacement reactions.

In a reactivity series, the metal which is present above in the series will easily displace a metal which is present below in the series.

We need to find an unknown metal which easily displaces cadmium metal from a solution which means that a metal is present above in the reactivity series and the same metal does not displace chromium which means that it is present below chromium. Hence, the unknown metal is iron.

Reactivity series is attached below.

8 0

4 years ago

Read 2 more answers