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show work and explain answer Question 1) part A : Use the data below to calculat

ID: 882586 • Letter: S

Question

show work and explain answer

Question 1) part A : Use the data below to calculate the heat of hydration of lithium chloride.

part B: Calculate the heat of hydration of sodium chloride.

2)What is the osmotic pressure of a solution made by dissolving 65.0 g of glucose, C6H12O6, in enough water to form 200.0 mL of solution at 43.0 C?

3)A certain reaction with an activation energy of 115 kJ/mol was run at 485 K and again at 505 K . What is the ratio of f at the higher temperature to f at the lower temperature? f505/f485=?

4)The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.30 kJ/mol and a frequency factor of A = 6.00×108M1s1 . The reaction is believed to be bimolecular:

NO(g)+F2(g)NOF(g)+F(g)

What is the rate constant at 657 C ?

5)In the first 14.0 s of the reaction, 1.7×102 mol of O2 is produced in a reaction vessel with a volume of 0.360 L . What is the average rate of the reaction over this time interval?

rate=_________M.s-1

part b:Predict the rate of change in the concentration of N2O over this time interval. In other words, what is [N2O]/t?

[N2O]/t=___________M.s-1

6)Consider the following data showing the initial rate of a reaction (Aproducts) at several different concentrations of A.

What is the order of the reaction?

n=_________?

Compound Lattice Energy (kJ/mol) Hsoln(kJ/mol) LiCl -834 -37.0 NaCl -769 +3.88

Explanation / Answer

1) the reation will be

a) Li+ + Cl - --> LiCl(s)   deltaH lattice =-834KJ /mole   .........equation 1

LiCl (s) + aq --> LiCl(aq) Delta H solution = -37 J / mole......equation 2

Li+ + Cl - --> LiCl (aq) Hydration

So Delta H hydration = Equation 1 + equation 2 = -834 + (-37 ) =-871 KJ / mol

b) Na+ + Cl - --> NaCl(s)   deltaH lattice = -769KJ /mole   .........equation 1

NaCl (s) + aq --> NaCl(aq) Delta H solution = 3.88J / mole......equation 2

Na+ + Cl - --> NaCl (aq) Hydration

So Delta H hydration = Equation 1 + equation 2 = - 769 + (+3.88 ) = - 765.12KJ / mol

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