The diatomic molecule H2 has an extremely low boiling point because only weak Vander waals forces are present between the atoms of hydrogen molecule. This weak inter-molecular forces are easily broken when heat is apply resulting in low boiling point.
Answer:
b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂
Explanation:
CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE
Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.
There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.
I hope it helps!
Answer:
Molecular formula = S₂O
₆
Explanation:
Given data:
Empirical formula = SO₃
Molecular formula = ?
Molecular mass = 152 g/mol
Solution:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 80 g/mol
n = 152 g/mol / 80 g/mol
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2(SO₃)
Molecular formula = S₂O
₆
Answer:
B: All of the choices!
Explanation:
Choice 1: Simply, there is no current without voltage, so it DOES cause current.
Choice 3: Voltage DOES push free electrons around a circuit. Without it, free electrons will move around between atoms, but randomly, so they wouldn't be much use.
Choice 4: Voltage IS measured in volts, so this option is true as well.
Choice 2: Voltage is all of those answers, so it is true! :D
Hope i helped! :]
Answer:
No effect will be observed since C is not included in the equilibrium expression.
Explanation:
Let's consider the following reaction.
CO₂(g) + C(graphite) ↔ 2 CO(g)
The equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.
Kc only includes gases and aqueous species (not liquids or solids).
![Kc=\frac{[CO]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCO%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
As we can see, C is a solid and is not included in the equilibrium expression, so adding some C would have no effect on it.
