Answer:
5.34 L
Explanation:
1.06×6.46×273/350.2×1
almost postive this is correct
Answer:
(a) The normal freezing point of water (J·K−1·mol−1) is
(b) The normal boiling point of water (J·K−1·mol−1) is 
(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole
Explanation:
Lets calculate
(a) - General equation -
=
= 
→ phases
ΔH → enthalpy of transition
T → temperature transition
=
=
(
is the enthalpy of fusion of water)
= 
(b) 
=
(
is the enthalpy of vaporization)
= 
(c)
=
°
°
=
°
°![C)]](https://tex.z-dn.net/?f=C%29%5D)
ΔT
°
°

= 109J/mole
Answer:
0.0084
Explanation:
The mole fraction of BaCl₂ (X) is calculated as follows:
X = moles BaCl₂/total moles of solution
Given:
moles of BaCl₂ = 0.400 moles
mass of water = 850.0 g
We have to convert the mass of water to moles, by using the molecular weight of water (Mw):
Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol
moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol
The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):
total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol
Finally, we calculate the mole fraction:
X = 0.400 mol/47.6 mol = 0.0084
Na is a cation (positive) and F is a anion (negative) this is an ionic bond. In a ionic bond the cation loses its valence electrons to give them to the anion. In this case Na will give its valence electron to F so F can complete its octet:)