Answer:
Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Explanation:
The equilibriums that take place are:
Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.
For Cu⁺²:
2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
2.2x10⁻²⁰ = 0.25 M*[OH⁻]²
[OH⁻] = 2.97x10⁻¹⁰ M
For Co⁺²:
1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²
[OH⁻] = 7.21x10⁻⁸ M
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Element with an atomic number of 58 is actually Cerium, so the symbol should be Ce, not Co because that is Cobalt which has an atomic number of 27. With that being said, the notation for isotopes is the symbol of the element with a superscript and a subscript that are aligned. The superscript represents the mass number.
Mass number = protons + neutrons = 58 + 33 = 91
The subscript is the atomic number which is 58. The notation is written in the picture attached.
Answer:
100.52
Explanation:
from the ideal gas equation PV=nRT
for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.
So n i.e no of moles will also be constant.
no of moles of Ar=3.224/40=0.0806
no of moles of unknown gas=0.0806
molecular wt of unknown gas=8.102/0.0806=100.52
Answer:
1) The bubbles will grow, and more may appear.
2)Can A will make a louder and stronger fizz than can B.
Explanation:
When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.
Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.
Answer:
n = 0.26 mol.
Explanation:
Given,
Pressure, P = 99.7 kPa = 1 atm
where 101.325 kPa = 1 atm
P = 0.984 atm
Temperature, T = 297 K
Volume = 6.452 L
Now, using ideal gas equation
PV = n RT
0.984 x 6.452 = n x 0.08206 x 297
n = 0.26 mol.
