Answer:
N3+ is ion of nitrogen is smaller
The second volume : V₂= 0.922 L
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Further explanation
</h3><h3>Given
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7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
Answer: from the Zn anode to the Cu cathode
Justification:
1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)
2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).
3) Then, you can already tell that electrons go from Zn to Cu.
4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.
So you get that the electrons flow from the anode (Zn) to the cathode (Cu).
Always oxidation occurs at the anode, and reduction occurs at the cathode.
Answer:
Explanation:
We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
Mᵣ: 44.01
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 1.5
1. Calculate the moles of CO₂
The molar ratio is 3 mol CO₂:1 mol C₃H₈
2. Calculate the mass of CO₂.
