Answer:
ΔH = 57.04 Kj/mole H₂O
Explanation:
60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)
=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)
=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)
=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat
ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O
=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)
ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2And that is the balanced chemical equation for the reaction shown in the figured attached.I hope this is useful for you..
The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.
Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.
We recognise that
h * 2 = PT - P * h * 20 = 1atm - 0.037atm
= 0.963 atm
1 * h * 2 = Ph * 2V / R * T
= 0.963 atm x 0.520 L / 0.0821 L atm/
molK * 301
= 0.02 mol h2
= 0.02molZn
So 0.02 mol Zn x 65.39 g/mol
= 1.33 g Zn
Read more about zinc;
brainly.com/question/28880469
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