Answer:
Explanation:
E = (hc)/(λ)
E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /
(77.8x10^(-9)m)
E = 2.55 x 10^(-11) J
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Answer: Statements (A), and (C) are correct.
Explanation:
The statements that are true are as follows.
- Particles in a liquid need to move more slowly in order to freeze.
When a liquid freezes the molecules get attracted towards each other. This attraction of particles occurs slowly. Hence, this statement is true.
- Attractive forces between the particles in a liquid are broken when a liquid boils.
When temperature is raised, the molecules in a liquid gains kinetic energy and start to move quickly in random directions. As a result, liquid state changes to gaseous state. Hence, this statement is true.
If the attractive force between gas molecules have to be increased, they should be moving slower instead because moving faster does not help attracting molecules together.
Hence, the statement particles in gas move fast enough to make more attractive forces when the gas condenses is not true.
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
An exothermic reaction releases heat. An endothermic reaction absorbs heat. Burning gas releases heat so it would be exothermic. Acid and water react heating the beaker would be exothermic because it releases heat from the reaction. Hope this helps! ;)
