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tatuchka [14]
2 years ago
5

Standardization is the process of titrating a solution prepared from

Chemistry
1 answer:
Zolol [24]2 years ago
6 0

Answer:

Standardization is the process of titrating a solution prepared from a carefully measured mass of solid accurately determine the concentration of the titrant.

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Why is physical exercise often suggested if you experience an abundance of stress?
scoray [572]

Answer:

good question

Explanation:

probably because people like to say they know things and help people with "there stress" and think that there helping them but in reality there making it where they have more :)

4 0
3 years ago
Drag each tile to the correct image. Match each hydrocarbon class to its structure. carboxylic acid amine halocarbon alcohol
Sonja [21]

Answer:

1. Amine.

2. Alcohol.

3. Carboxylic Acid.

4. Halocarbon.

Explanation:

3 0
3 years ago
several dams on the colorado river in texas produce electricity through hydropower. what is an advantage of this type of energy
Ivanshal [37]

Explanation:

hydropower energy is:

-renewable

-clean (doesn't produce toxins or emit harmful gases into the atmosphere)

-environment friendly

-cheap once installed

3 0
2 years ago
A gas at 1.5 atm had pits ressure decreased to 0.50 atm producing a new volume of 750 ml what was its original volume
Vlada [557]

Answer:

The original volume of the gas is 0.001 mL

Explanation:

This easy excersise can be solved by the law for gases, about pressure and volume; the volume of a gas is inversely proportional to the pressure it exerts.

We can propose the rule by this formula:

P₁  / V₁ = P₂ / V₂

We replace data given: 1.50 atm / V₁ = 0.50 atm / 750 mL

As the rule says, that volume is inversely proportional, and the pressure was decreased, volume must be lower than 750 mL.

1.5atm / (0.5 atm / 750mL) = V₁

V₁ = 0.001 mL

6 0
3 years ago
0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

4 0
3 years ago
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