Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m
Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:
48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ
<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>
Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
Answer:
activity coefficient 
activity coefficient 
The change in pH in part A = 0.092
The change in pH in part B = 0.102
Explanation:
From the given information:
pH of HCl solution = 1.092
Activity of the pH solution [a] = 
[a] = 0.0809 M
Recall that [a] =
× C
where;
= activity coefficient
C = concentration
Making the activity coefficient the subject of the formula, we have:
![\gamma = \dfrac{[a]}{C}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20%5Cdfrac%7B%5Ba%5D%7D%7BC%7D)


B.
The pH of a solution of HCl and KCl = 2.102
[a] = 
[a] = 0.00791 M
activity coefficient 

C. The change in pH in part A = 1.091 - 1.0 = 0.092
The change in pH in part B = 2.102 -2.00 = 0.102
i think the greater the electric charge the atom decreases in size