The density of liquid and gaseous nitrogen and the volume of the liquid nitrogen are given. The fraction of air that is displaced in the room of given dimensions by the gaseous nitrogen is to be calculated. Concept introduction: The conversion of one liter into milliliter is done as, 1 L = 10 3 mL The mass is calculated by the formula, Mass = Density × Volume The volume is calculated by the formula, Volume = Mass Density The conversion of one millimeter into m 3 is done as, 1 mL = 10 − 6 m 3 To determine: The fraction of air in the room that is displaced by the gaseous nitrogen.
The density of liquid and gaseous nitrogen and the volume of the liquid nitrogen are given. The fraction of air that is displaced in the room of given dimensions by the gaseous nitrogen is to be calculated. Concept introduction: The conversion of one liter into milliliter is done as, 1 L = 10 3 mL The mass is calculated by the formula, Mass = Density × Volume The volume is calculated by the formula, Volume = Mass Density The conversion of one millimeter into m 3 is done as, 1 mL = 10 − 6 m 3 To determine: The fraction of air in the room that is displaced by the gaseous nitrogen.
The density of liquid and gaseous nitrogen and the volume of the liquid nitrogen are given. The fraction of air that is displaced in the room of given dimensions by the gaseous nitrogen is to be calculated.
Concept introduction: The conversion of one liter into milliliter is done as, 1L=103mL
The mass is calculated by the formula, Mass=Density×Volume
The volume is calculated by the formula,
Volume=MassDensity
The conversion of one millimeter into m3 is done as, 1mL=10−6m3
To determine: The fraction of air in the room that is displaced by the gaseous nitrogen.
Please do these questions within the SCH4U course please with full steps since I am still unsure how to format my answers! Thank you so much.
When two solutions, one of 0.1 M KCl (I) and the other of 0.1 M MCl (II), are brought into contact by a membrane. The cation M cannot cross the membrane. At equilibrium, x moles of K+ will have passed from solution (I) to (II). To maintain the neutrality of the two solutions, x moles of Cl- will also have to pass from I to II. Explain this equality: (0.1 - x)/x = (0.1 + x)/(0.1 - x)
Calculate the variation in the potential of the Pt/MnO4-, Mn2+ pair with pH, indicating the value of the standard potential. Data: E0 = 1.12.
Chapter 1 Solutions
Principles of Chemistry: A Molecular Approach, Books a la Carte Edition; Modified Mastering Chemistry with Pearson eText - ValuePack Access Card - Chemistry: A Molecular Approach (3rd Edition)
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