A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Answer: Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass.
Explanation: Hope this helped! :)
Answer:
A
Explanation:
There are three states of mater; solid liquid and gas. The sold state is the difficult to compress while the gaseous state is quite easy to compress.
A gas is easily compressed because the particles in a gas are far apart from each other. A solid is difficult to compress because the particles of a solid are close together. From all the above statements, it is easily deducible that the compressibility property of a substance in a particular state of matter depends on the proximity of the particles to each other, hence the answer above.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
