ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants)
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) )
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 )
ΔH(reaction) = -283 - ΔH(PbO(s))
-131.4 = -283 -ΔH(PbO(s))
ΔH(PbO(s)) = -151.6 kJ
So, the best answer is A.
Answer:
Do to half of the mnairals this can not be made into a lab there is an error
Explanation:
Answer:
The number of solute particles increases, and the boiling point increases.
Explanation:
- It is known from colligative properties that adding solute to the solvent will cause elevation of boiling point.
- Elevation of boiling point (ΔTb) can be expressed as:
<em>ΔTb = Kb.m,</em>
where, Kb molal boiling point elevation constant.
m is the molal concentration of solute.
- Adding more sodium chloride to the solution:
will increase the number of solute particles and also will increase the molal concentration of NaCl solute.
<em>∵ ΔTb ∝ m.</em>
- So, the boiling point increases.
- Thus, the right choice is:
<em>The number of solute particles increases,</em>
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Using a more concentrated HCl solution and Crushing the CaCO₃ into a fine powder makes the reaction to occur at a faster rate.
<u>Explanation:</u>
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(aq) + CO₂(g)
When calcium carbonate reacts with hydrochloric acid, it gives out carbon-dioxide in the form of bubbles and there is a formation of calcium chloride in aqueous medium.
The rate of the reaction can be increased by
- Using a more concentrated HCl solution
- Crushing the CaCO₃ into a fine powder
When concentrated acid is used instead of dilute acid then the reaction will occur at a faster rate.
When CaCO₃ is crushed into a fine powder then the surface area will increases thereby increasing the rate of the reaction.
Answer:
B = CHCl2 + Cl2 --> CHCl3 + Cl
Explanation:
Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.
Free radical chlorination is divided into 3 steps which are:
The initiation step
The propagation step
The termination step
So in reference to the question, propagation step involves two steps.
The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.
The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.
You would find in the attachment the 2 step mechanism.
