As the temperature increases, the solubility of the solute in the liquid also increases. This is due to the fact that the increase in energy allows the liquid to more effectively break up the solute. The additoin of energy also shifts the equilibrium of the reation to the right since it takes energy to dissolve most things and you are adding more of it (this is explained with Le Chatlier principles).
I hope this helps and also I assumed that your question involved the solubility of an ionic substance in a solvent like water. If that was not your question feel free to say so in the comments so that I can answer your actually question.
<span>The pH scale goes from 0-14. 0-6.9 is acidic, 7 is neutral and 7.1-14 is basic</span>
<h2><u>
Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
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Explanation:</u></h2>
<u></u>
PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
Answer:
0.571 mol
Explanation:
Given data:
Number of moles of NaHCO₃ = 0.571 mol
Number of moles of CO₂ produced = ?
Solution:
Chemical equation:
NaHCO₃ + C₃H₆O₃ → CO₂ + C₃H₅NaO₃ + H₂O
Now we will compare the moles of CO₂ with NaHCO₃ from balance chemical equation.
NaHCO₃ : CO₂
1 : 1
0.571 : 0.571
So number of moles of CO₂ produced are 0.571.
Answer:
it’s mass was greater than when it started
Explanation:
When a metal is coated with another metal, the plating metal deposits on the plated metal. Usually, the plating metal functions as the anode while the plated metal functions on the cathode. The anode metal is oxidized and reduced at the cathode and become deposited on the cathode material. This increases the mass of the cathode. Hence the mass of the silver/gold product is greater than the mass of silver at the beginning of the electroplating process.
