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>.
It is an example of a molecule
Answer : The temperature of neon gas will be, 221.0 K
Explanation :
To calculate the temperature of neon gas we are using ideal gas equation.
where,
P = pressure of neon gas = 96.7 kPa = 0.955 atm
Conversion used : (1 atm = 101.3 kPa)
V = volume of neon gas = 9.50 L
T = temperature of neon gas = ?
R = gas constant = 0.0821 L.atm/mole.K
w = mass of neon gas = 10.00 g
M = molar mass of neon gas = 20 g/mole
Now put all the given values in the ideal gas equation, we get:
Therefore, the temperature of neon gas will be, 221.0 K