Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
1.77 so A. I don’t need to explain this just helping out haha
The freezing point depression can be explained in terms of vapor pressure adding solvent will decrease molecules
Answer:
to the right (products side)
Explanation:
The equilibrium constant K describes the ratio between the concentration of products and reactants at equilibrium. For a general reaction:
a A + b B → c C + d D
The equilibrium constant expression is:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%20%7D)
A low value of K indicates that the concentration of products (C and D) is low in relation with the concentration of reactants (A and B).
Conversely, a high value of K indicated that the concentration of products is high compared with the concentration of reactants.
Since K = 6.4 × 10⁹ is a high value, the concentration of products is higher than the concentration of reactants at equilibrium. Thus, the position of the equilibrium is favored to the right.
