Answer:
- <u><em>Sodium chloride</em></u>
Explanation:
The attached graph with a green and a red arrow facilitates the understanding of this explanation.
To read the <em>solubility </em>on the <em>graph</em>, you can start with the temperature, on the x-axis.
The red vertical arrow shows how, departing from the <em>40ºC temperature</em> on the x-axis, you intersect the<em> solutibility curve </em>of sodium chloride at a height (y-axis) corresponding to <em>60 g/100cm³ of water</em> (follow the green horizontal arrow).
Hence, <em>sodium chloride is the salt that can dissolve at a concentration of about 60g/100cm³ of water at 40ºC.</em>
Answer: Kb = 3.15 × 10 ⁻⁴
Explanation:
This is how you calculate Kb for this reaction.
1) Equilibrium equation:
CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻
2) Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂] ↔ all the spieces in equilibrium
3) From the stoichiometry [CH₃NH₃⁺] = [OH⁻]
Then, Kb = [OH⁻] [OH⁻] / [CH₃NH₂] = [OH⁻]² / [CH₃NH₂]
4) You get [OH⁻] from the pH in this way:
pOH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 11.40 = 2.60
pOH = - log [OH⁻] = 2.60 ⇒ [OH⁻] = 10^(-2.6) = 0.002512
5) [CH₃NH₂] in equilibrium is given: 0.0200M
6) Now compute:
Kb = (0.002512)² / 0.0200 = 3.15 × 10 ⁻⁴
Answer:
106.6 °C
Step-by-step explanation:
The formula for <em>boiling point elevation</em> ΔTb is
ΔTb = iKb·b
where
i = the van't Hoff i# factor
Kb = the molal boiling point elevation constant
b = the molal concentration of the solution
=====
<em>Data
</em>
i = 2, because 1 mol of NaCl gives 2 mol of ions in solution.
Kb = 0.51 °C·mol·kg⁻¹
b = 1.00/0.155
b = 6.452 mol·kg⁻¹
=====
<em>Calculations
</em>
ΔTb = 2 × 0.51 × 6.452
ΔTb = 6.58°C
Tb = Tb° + ΔTb
Tb = 100 + 6.58
Tb = 106.6 °C
stronger than those in a liquid
Answer:
radiation
Explanation:
Radiation is light. the lamps above are producing heat through the light.
