Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
![K_c=\frac{[H_2][Br_2]}{[HBr]^2} \\\\=\frac{(0.395)(0.395)}{(1.361)^2} \\\\=\frac{0.156025}{1.852321} \\\\=0.084](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%280.395%29%280.395%29%7D%7B%281.361%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B0.156025%7D%7B1.852321%7D%20%5C%5C%5C%5C%3D0.084)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Answer:
-4.59°C
Explanation:
Let's see the formula for freezing point depression.
ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Kf = Freezing constant. For water if 1.86°C/m
m = molality (moles of solute in 1kg of solvent)
i = Van't Hoff factor.
0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9
Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)
Freezing T° of solution = - (1.86°C/m . 1.30m . 1.9) → -4.59°C
NaCl → Na⁺ + Cl⁻
Answer:
False
Explanation:
The Avogadro's number is not used to determine the number of subatomic particles in an atom.
Subatomic particles of an atom are the protons, neutrons and electrons.
The protons are the positively charged particles in an atom
Neutrons do not carry any charges
Electrons carry negative charges.
The number of protons, neutrons and electrons in an atom are experimentally determine using spectrometric techniques.
Hey! Your answer would be the appearance and disappearance of group of organisms
. Hope this helps!
