It means that <span>the cell loses most of its water from osmosis when put in a hypertonic.
Hope that helps!</span>
Data validation → editing data → entry data → tabulation
This enables better integration, consumption and analysis of larger datasets using advanced business intelligence with analytics solutions.
<h3>What is data preparation ?</h3>
Data collection, combination, structure, and organisation are all steps in the process of preparing data for use in business intelligence (BI), analytics, and data visualisation applications.
- Data prep is a common colloquial term for data preparation. It is also referred to as "data wrangling," while other professionals use that phrase more specifically to mean "cleaning, structuring, and transforming data." This usage distinguishes data wrangling from the data pretreatment stage.
Learn more about Data preparation here:
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<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:
Or,
where,
= 
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:
Hence, the freezing point of solution is 2.6°C
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
The answer is D, because "Animals without a backbone are invertebrates."
