To solve this we must be knowing each and every concept related to experiment. Therefore, the correct option is option B among all the given options.
An experiment is a qualitative approach used in the scientific process to settle disputes between competing products or hypotheses. Additionally, experiments are used by researchers to confirm or refute new or existing theories.
A student designs an experiment where she will put the same amount of air into three different balloons. One balloon will be submerged in ice water, one balloon will be carefully heated with a hair dryer, and the last balloon will be left alone. After making the changes, the student then will then measure the diameter of each balloon. The dependent variable is the diameter of the balloon.
Therefore, the correct option is option B.
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Answer:
B
Explanation:
B. solute
C. solvator
D. emulsifier
Answer : The correct option is, (A) solvent
Explanation :
Solvent : It is present in large amount. It is a type of substance in which the amount of solute is dissolved.
Solute : It is present in smaller amount. It is a substance that is dissolved in another substance.
Emulsifier : It is a molecule that has two ends, one is water loving end and another is oil loving end. That means it is a substance which stabilizes the emulsion.
Hence, the term for the dissolving medium in a solution is, solvent.
The term for the dissolving medium in a solution is "solvent." It is the component that can dissolve other substances (solute) to form a homogeneous mixture, known as a solution.
The term for the dissolving medium in a solution is "solvent." In a solution, a solvent is the component that exists in the larger quantity and has the ability to dissolve other substances, known as solutes. This process occurs due to the solvent's molecular structure, which allows it to interact with and surround the solute particles, breaking their intermolecular forces and dispersing them evenly throughout the solvent.
Solvents can be liquids (such as water in aqueous solutions), solids (as in solid-state solutions), or even gases (in gas-phase solutions). The choice of solvent depends on the specific application and the nature of the solute. For instance, in chemistry labs, common solvents include water, ethanol, acetone, and more, each chosen based on their compatibility with the substances being studied.
The ability of a solvent to dissolve a solute is influenced by various factors, including temperature, pressure, and the nature of the solute and solvent molecules. The dissolution process is often described in terms of solvation, where solvent molecules surround and stabilize solute particles through electrostatic interactions.
In summary, the solvent is a crucial component of a solution, responsible for dissolving and dispersing solute particles, leading to the formation of a homogeneous mixture. This concept is foundational in chemistry and plays a fundamental role in various scientific, industrial, and everyday applications.
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i) 737 K = 463.85 °C
ii) -265 °F = -165 °C
Converting 737 K into °C :
As we know Absolute zero (0K) is equivalent to -273. 15 °C. Hence,
K = °C + 273.15
Or,
°C = K - 273.15
Putting value of K,
°C = 737 - 273.15
°C = 463.85 °C
Converting -265 °F into °C :
As,
°F = °C × 1.8 + 32
So,
°C = °F - 32 ÷ 1.8
Putting value of °F,
°C = -265 - 32 ÷ 1.8
°C = -165 °C
Answer:
Freezing T° of solution = - 48.12°C
Explanation:
The colligative which has to be used for this case is the freezing point depresison ( ΔT = Kf . m )
ΔT = Freezing T° of solvent - Freezing T° of solution
Kf = Crysocopic constant
m = molality (mol/kg)
We determine the molality (moles of solute in 1kg of solvent)
We convert the mass of CCl₄ from g to kg → 500 g . 1kg / 1000g = 0.5kg
0.42 mol of hexane / 0.5 kg of CCl₄ = 0.84 mol/kg
Let's replace data: -22.92°C - Freezing T° of solution = 30°C/m . 0.84 m
Freezing T° of solution = - ( 30°C/m . 0.84 m + 22.92°C) → - 48.12°C
Answer:
moist air is less dense than dry air because water has a lower molecular weight than nitrogen and oxygen
Answer:
Water vapor is a relatively light gas when compared to diatomic Oxygen and diatomic Nitrogen. Thus, when water vapor increases, the amount of Oxygen and Nitrogen decrease per unit volume and thus density decreases because mass is decreasing.