The answer is the first one
Answer:
0.13 M ( 2 s.f)
Explanation:
2Cl2O5 (g)-->2Cl2(g) +5O2 (g)
rate= (17.4 M -1 .s -1 ) [Cl2O5]2
From the rte above, we can tell that our rate constant (k) = 17.4 M -1 .s -1
The units of k tells us this is a second order reaction.
Initial Concentration [A]o = 1.46M
Final Concentration [A] = ?
Time = 0.400s
The integrated rate law for second order reactions is given as;
1 / [A] = (1 / [A]o) + kt
1 / [A] = [ (1/ 1.46) + (17.4 * 0.4) ]
1 / [A] = 0.6849 + 6.96
1 / [A] = 7.6496
[A] = 1 / 7.6496
[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)
Answer:
In a parallel circuit, current divides through resistors and current might be different depending upon the resistor and all resistors have the same potential difference. Therefore, if a parallel resistor was removed then the total resistance of the circuit will increase.
Answer:
Explanation:
Mineral
The naturally occurring mineral anglesite, PbSO4, occurs as an oxidation product of primary lead sulfide ore, galena.
Basic and hydrogen lead sulfates
A number of lead basic sulfates are known: PbSO4·PbO; PbSO4·2PbO; PbSO4·3PbO; PbSO4·4PbO. They are used in manufacturing of active paste for lead acid batteries. A related mineral is leadhillite, 2PbCO3·PbSO4·Pb(OH)2.
At high concentration of sulfuric acid (>80%), lead hydrogensulfate, Pb(HSO4)2, forms.[4]
Chemical properties
Lead(II) sulfate can be dissolved in concentrated HNO3, HCl, H2SO4 producing acidic salts or complex compounds, and in concentrated alkali giving soluble tetrahydroxidoplumbate(II) [Pb(OH)4]2− complexes.
PbSO4(s) + H2SO4(l) ⇌ Pb(HSO4)2(aq)
PbSO4(s) + 4NaOH(aq) → Na2[Pb(OH)4](aq) + Na2SO4(aq)
Lead(II) sulfate decomposes when heated above 1000 °C:
PbSO4(s) → PbO(s) + SO3(g)
