Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)

Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α

Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)



Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)
Answer:
AgNO3
Explanation:
Silver nitrate (AgNO3) is an inorganic compound. Organic compounds usually have 1+ C.
Answer:
(40 g O) / (15.99943 g O/mol) x (1 mol CaCO3 / 3 mol O) x (100.0875 g CaCO3/mol) =
83 g CaCO3
So answer D), although three significant digits are not justified.
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