Answer:
1) The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.
2) The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt = 0.045 mol/L.s.
Explanation:
<em>3NO(g) → N₂O(g) + NO₂(g).</em>
The rate of the reaction = -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt = Δ[NO₂]/Δt.
Given that: Δ[N₂O]/Δt = 0.015 mol/L.s.
<em>1) The rate of the overall reaction is?</em>
The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.
<em>2) The rate of change for NO is?</em>
The rate of change for NO = - Δ[NO]/Δt.
∵ -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt.
<em>∴ The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt </em>= 3(0.015 mol/L.s) = <em>0.045 mol/L.s.</em>
Pneumonia is an infection of the lungs caused by fungi, bacteria, ... These conditions include asthma, cystic fibrosis, diabetes, and heart failure.
Answer:
im pretty sure its b. sand and iron
ik that magnetism easily separates iron
The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
