Answers:
1st: 189.6 g/mol
2nd: 0.1357 L
3rd: 1.41 M
Explanation:
Finding Molar Mass:
SnCl2 = <u>Tin(II) Chloride</u>
Tin has a molar mass of <u>118.71 g/mol</u>
Chloride has a molar mass of <u>35.453 g/mol</u>
Chloride*2 = <u>70.906</u>
<u>118.71 + 70.906 ≈ 189.6 g/mol</u>
Finding Liters of Solution:
L = mL/1000
135.7 mL / 1000 = <u>0.1357</u>
Finding Molarity:
molarity = <u>moles of solute / liters of solution</u>
M = (36.4g / 189.6g) / 0.1357 L = <u>1.41 M</u>
Hope this helped ;)
<span>1 mole of benzene (78g) requires 30.8 kJ/ of heat, so 11.5g will need ..... (it's a proportion calculation.) Temperature does not change at BPt and is not relevant if the temp of the liquid is already at the BPt
ne definition of entropy is qrev/T, where qrev is the heat added in reversible operation (for complicated reasons pertaining to heat as a path function) and T is the temperature at which this is done.
Phase changes are particularly good examples for calculations of changes in entropy, since temperature will not change will the bonds of a state are being broken.
The calculations required boils down to:
1) finding the moles of benzene given from molar mass.
2) multiplying that moles by the heat of vaporization.
3)diving the heat energy required by the temperature of boiling point.</span>
Answer:
pH= 8.45
Explanation:
when working with strong accids pH = -log(Concentration)
so -log(3.58e-9) = 8.446
<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol
