Jaką maksymalną ilość gramów azotanu (V) potasu można rozpuscic w 300g wody w temperaturze 90 C
1 answer:
Answer:
- <u>609g </u>
<u></u>
Explanation:
The translated question is:
<em>What maximum amount of grams of potassium nitrate (V) can be dissolved in 300g of water at 90 °C</em>
<em></em>
<h2>Solution</h2><em></em>
To answer the question you need to consultate the solubiity information for potassium nitrate (V), KNO₃.
The attached table contains the solutibility table for KNO₃ at different temperatures.
At 90ºC it is 203g / 100g water.
Then, to calculate the <em>maximum amount of grams of potassium nitrate (V) that can be dissolved in 300g of water at 90 °C</em>, just multiply by the amount of water:
- 203g / 100g water × 300 g water = 609g ← answer
You might be interested in
Answer:
a) galvanic cell
b)electrolytic cell
c) i) K=6.27x10'34
ΔG°=198790 J
ii) K=3.58x10'-34
ΔG°= 191070 J
d) E°=0.278 v
ΔG°= -26827 J
Explanation:
a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".
The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.
b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.
c) i) look image attached
ii) k = look image attached
ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)
ΔG° =-191070
d) E°= 0.0592 v/n x lg K
E°= 0.0592V / 1 X log 5.0X10'4
E°= 0.278 v
ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v
ΔG° = -26827 J
