Answer: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!
group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1
we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!
Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.
Answer:
300000Pa or 3×10^5 Pa
Explanation:
Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.
Using Boyle's law
P1V1 = P2V2
P1 is the initial pressure = 1.5×10^5Pa
V1 is the initial volume = 0.08m3
P2 is the final pressure (required)
V2 is the final volume = 0.04 m3
From the formula, P2 = P1V1/V2
P2 = 1.5×10^5 × 0.08 ÷ 0.04
= 300000Pa or 3×10^5 Pa.
Answer:
2.772 seconds
Explanation:
Given that;
t1/2 = 0.693/k
Where;
t1/2 = half life of the reaction
k= rate constant
Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant
t1/2 = 0.693/2.5 x 10-1 s-1
t1/2= 2.772 seconds
