Answer:
1. d. It's water
2. a. It does not always retain the properties of the substances that make it up
Explanation:
They are atoms like hydrogen oxygen or nitrogen
Answer:
First, let's determine how many moles of oxygen we have.
Atomic weight oxygen = 15.999
Molar mass O2 = 2*15.999 = 31.998 g/mol
We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml
Since the density is 1.149 g/mol,
we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2
Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol
Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K)
Now take the formula and solve for V, then substitute the known values and solve.
PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L
So the volume (rounded to 3 significant figures) will be 134 ml.
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
