2K(s)+2H2O(l)→2K+(aq)+2OH−(aq)+H2(g).
with the equation above,
part C
Assume that a chunk of potassium weighing 8.60 g is dropped into 400.0 g of water at 29.0 ∘C. What is the final temperature of the water if all the heat released is used to warm the water?
PART D
What is the molarity of the KOH solution prepared in part (c)?
Following is the - complete Answer -&- Explanation: for the given: Question: in....typed format...
Answer:
Explanation:
Following is the complete Explanation: for the above: Answer.
We know the following: molar masses:
We know the following formula: for determining the value of standard enthalpy of reaction.
Horxn = vp x( Hof[ products ) ] - vr x( Hof [ reactants ) ] -------------------Equation - 2
Where:
We know, the following standard enthalpy of formation: values:
Therefore: we can plug in the values in Equation - 2, above to find the value of the standard enthalpy of reaction:
Horxn..
We know: the number of moles of potassium = ( 8.60 g ) / ( 40.0 g/mol ) = 0.215 mol ( moles )
From: Equation - 1: we know: the following molar ratio:
molar ratio: K : H2O : K+ : OH - : H2 = 1 : 1 : 1 : 1 : 0.5 ( i.e. by dividing Equation - 1: by 2.0 )
Therefore: if we have: 0.215 moles, of potassium ( K ) ; we will have the following moles of the compounds:
We have the following: Equation - 2; after we plug in the values:
Horxn = [ (0.215 mol ) x ( Hof [ K+ (aq) ] ) + (0.215 mol ) x ( Hof [ OH - (aq) ]) + (0.1075 mol ) x ( Hof [ H2 (g) ] ) ] - [ (0.215 mol ) x ( Hof [ K(s) ] ) + ( 0.215 mol ) x ( Hof [ H2O (l) ] )
Horxn = [ (0.215 mol ) x ( - 251.2 kJ/mol ) + (0.215 mol ) x ( - 229.9 kJ /mol ) + (0.1075 mol ) x (0.0 kJ/mol) ] - [ (0.215 mol ) x ( 0.0 kJ/mol ) + ( 0.215 mol ) x ( - 285.8 kJ /mol ) ]
Horxn = - 41.88 kJ ( i.e. the minus sign indicates , release of heat )
We know the above: heat energy: will heat up the given amount of water, in which the given mass of potassium: was dropped into.
Therefore: ..
Therefore: we can:say that the amount of heat absorbed by water = Horxn, and using the Equation of Heat Transfer: we will get the following:
mw x Cw x T = ( 400.0 g ) x ( 4.184 J / g. oC ) x ( Tf - 29.0 ) oC = 41880 J
( Tf - 29.0 ) oC = 25.02 oC, therefore: ...
Final Temperature: Tf = ( 25.02 + 29.0 ) = 54.02 oC
Final temperature of water: Tf = 54.02 oC
Since, as we know: 0.215 moles Of KOH: will remain dissolved in 400.0 g or 400.0 mL of Water, we can say:
400.0 mL of water contains: 0.215 moles Of KOH:
1000.0 mL of water will contain: [( 0.215 mol ) / ( 400.0 mL ) ] x ( 1000 mL ) = 0.5375 moles Of KOH
Therefore:
Molarity of 0.215 moles Of KOH: = 0.5375 M ( mol/L )
2K(s)+2H2O(l)→2K+(aq)+2OH−(aq)+H2(g). with the equation above, part C Assume that a chunk of potassium weighing 8.60 g...
PART C Assume that a chunk of potassium weighing 8.60 g is dropped into 400.0 g of water at 29.0 ∘C. What is the final temperature of the water if all the heat released is used to warm the water?
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