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:
2 CH2 + 3 O2 = 2 CO2 + 2 H2O
Explanation:
This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.
Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.
When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.
However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.
Answer:
By weight they have the same mass, but the number of atoms is different
Explanation:
Answer:
5 000 000 (5 million atoms)
Explanation:
Let us assume that a vanadium atom has a spherical shape.
diameter of a sphere = 2 x radius of the sphere
Thus,
Radius of a vanadium atom = 130 pm
= 130 x
m
The diameter of a vanadium atom = 2 x radius
= 2 x 130 x
= 260 x
m
Given a distance of 1.30 mm = 1.30 x
m,
The number of vanadium atoms required to span the distance = 
= 5000000
Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.