Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= 
(At standard temperature)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Answer:
2NaOH (aq) + CaCl2 (aq) -> 2NaCl(aq) + Ca(OH)2(s)
Formula of precipitate: Ca(OH)2 <em>(s)</em>
Explanation:
First, we do the double replacement reaction to determine our chemical equation between the reactants and products. Once we have our products, with a solubility chart (I added one below) we can determine which of the products is soluble or insoluble.
In this case NaCl is soluble or aqueous (meaning it can dissolve in water) and Ca(OH)2 is insoluble (meaning that when the reactions takes place, these two will form a solid/precipitate)
Of the many processes involved in the water cycle, the most important are evaporation, transpiration, condensation, precipitation, and runoff. Although the total amount of water within the cycle remains essentially constant, its distribution among the various processes is continually changing.
Answer:
The answer is "Each student will get a glass of water and drop the bead into it but the beads float 0.6 g / cm3 and slip down to 1.2 g / cm3
".
Explanation:
One's masses would've been dissimilar, even though their width and concentrations were also equal. Whenever the type-A mass is m, then the type-B mass is 2 m. One should measure then, therefore.
Water has a 1g / cm^3 density. Although Type A is higher than air, Type B is much more compact. it means will float if they place it in water type-A where type-B sinks.
