Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
Answer:
Frecuency = 5,83x10⁻⁷ Hz
Explanation:
The equation that connects wavelenght and frequency is given by:
λ = c/ν
λ=wavelenght (expressed in lenght´s units)
c= speed of light (3x10⁸ m/sec)
ν=frequency (expressed in units of time⁻¹ or Herzt)
In our case, λ=5,14x10⁻⁷ m , so replacing in our previous formula, this gives us the final result of ν (frequency for green light) of 5,83x10¹⁴ Hz (or Herzt)
Answer:
CH₄
Explanation:
CH₃OH has hydrogen bonding due to the OH group present
NH₃ also has hydrogen bonding due to the NH bonds
H₂S has dipole-dipole forces present due to the polar SH bonds
HCl also has dipole-dipole forces due to the polar HCl bond
Answer:
The amount of NO₂ that can be produced 8.533 g
Explanation:
According to question
2 NO(g) + O₂(g) → 2 NO₂(g)
Given
Moles of nitrogen monoxide = 0.377
Moles of oxygen = 0.278
Since 'NO' is the limiting reagent according to this ratio.
According to equation
2 moles NO reacts to form 2 moles NO₂
So, 0.1855 moles NO give = 0.1855 moles of NO₂
Mass of 1 mole NO₂ = 46 g/mole
Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g
Answer:
a . knowledge and existing theories .
b. falsifiable scope
