Answer:
Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.
Explanation:
Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:
Boiling Point
Density
State of Matter
Color
Melting Point
Odor
Temperature
Refractive Index
Luster
Hardness
Ductility
Malleability
Answer: Marcia made a more concentrated salt solution.
Explanation:
Marcia used 1 kg of salt, which the exercise tells us is 2.2 lb.
She dissolved it in 3L of water, so we have to find out how much that is in pints. We do that using cross multiplication.
If 1L equals 2.11 pints, then
xpints / 3L = 2.11pints / 1L, so
xpints = (3L . 2.11pints) / 1L = 6.33 pints
Now we know she dissolved 2.2lb in 6.33 pints of water. We use cross multiplication again to find out how much salt she would have in 10 pints, so we can know how concentrated it is compared to Bobby's solution, which is 10 pints.
xlb / 10pints = 2.2lb / 6.33pints
xlb = (10pints . 2.2lb) / 6.33pints = 3.47lb
So Marcia has a concentration of 3.47lb per 10 pints, whereas Bobby only has 3lb per 10 pints.
Answer : The mass of the water molecule is 4.5 times greater than the mass of the helium atom.
Explanation :
Assumption : The number of water molecules is equal to the number of helium atoms
Given : The mass of water = 4.5 × The mass of helium ........(1)
The mass of Water = Mass of 1 water molecule × Number of water molecule
The mass of Helium = Mass of 1 helium atom × Number of helium atom
Now these two masses expression put in the equation (1), we get
Mass of 1 water molecule × Number of water molecule = 4.5 × Mass of 1 helium atom × Number of helium atom
As per assumption, the number of water molecules is equal to the number of helium atoms. The relation between the mass of water molecule and the mass of helium atom is,
Mass of water molecule = 4.5 × Mass of helium atom
Therefore, the mass of the water molecule is 4.5 times greater than the mass of the helium atom.
AH1 = m * c1 * AT1 calculate this for ice (-25C to 0C) AH2 = AHfus(1 mole)=6.01 kJ = 6010 J AH3 = m *c3 * AT3 calculat this for water (0C to 100C) AH4 = AHvap(1mole)=40.67 kJ = 40670 J AH5= m * c5 * AT5 calculate this for steam (100C to 125C)
Sum ---- AH1+AH2+AH3+AH4+AH5
Data m=18g (1mole water)
c1=specific heat ice= 2.09 J/g K c3=specific heat water= 4.18 J/g K c5=specific heat steam= 1.84 J/g K
AT = (Tend - Tinitial) as this is a difference between temperatures it doesn't matter the units Celsius or Kelvin. Kelvin (K)=Celsius (C)+273.15
AT1 = 0C - (-25C)= 25C= 273.15K - 248.15K= 25K AT3= 100C - 0C = 100C= 100K AT5= 125C - 100C= 25C=25K
