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noname [10]
3 years ago
8

When does phenolphalien turn pink

Chemistry
1 answer:
rewona [7]3 years ago
3 0

Answer : Phenolphalien turn pink in the presence of a base.

Phenolphthalien : It is an indicator which is used in the acid-base titration.

In acidic medium, the phenolphthalien is colorless and the pH will be less than 7.

In basic medium, the phenolphthalien is pink in color and the pH will be greater than 7.

Phenolphthalien is slightly soluble in the water and highly soluble in alcohols.

Hence, the phenolphalien turn pink in the presence of a base.

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Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar
kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S_{Generation = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S_{Generation = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S_{Generation = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S_{Generation = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0
2 years ago
Consider the following Reaction.
jeyben [28]

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol


number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 =>  0.5246 moles

Molar mass of H3PO4 =  97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:


3 Mg(OH)2 + 2 H3PO4 --->  Mg3(PO4)2 + 6 H2O


3 mol of Mg(OH)2 reacts with 2 mol of H3PO4  ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation  .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation  :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2


= (1/3)*0.5246


= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass


= 0.1749 *  262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%


Hope that helps!




3 0
3 years ago
What is a good title for this chart?
Elina [12.6K]

Answer:

pH of the acid

Explanation:

8 0
3 years ago
Explanation about the combining powers of element chemistry
stiv31 [10]

Answer:

<em>Valency is the combining power of an element. Elements in the same group of the periodic table have the same valency. The valency of an element is related to how many electrons are in the outer shell. The noble gases have the valency 0 as they do not usually combine with other elements.</em>

Explanation:

HOPE IT HELPS

5 0
3 years ago
Which of the following is NOT a characteristic of stars? A) Size B) temperature C) texture D) color
Lynna [10]

Answer:

C

Explanation:

Characteristics used to classify stars include color, temperature, size, composition, and brightness.

7 0
3 years ago
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