B - two hydrogen atoms and one oxygen atom
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
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CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
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1) Ca(OH)2 should be reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s)
we are going to take as
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>
</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)
Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
You determine the correct number of significant figure of a measurement in a graduated cylinder by looking at the smallest division on the graduated cylinder. If the division is divided up to the ones place, you can still read the half of that division or the .50 measurement. So, you can have until the tenths place in your measurement.
Answer:
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Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
