Answer:
0.055g/mL
Explanation:
Data obtained from the question include:
Molar Mass of the gass sample = 71g/mol
Volume of the gas sample = 1300 mL
Density =?
The density of a substance is simply mass per unit volume. It is represented mathematically as:
Density = Mass /volume.
With the above equation, we can easily obtain the density of sample of gas as illustrated below:
Density = 71g / 1300 mL
Density = 0.055g/mL
Therefore, the density of the gas sample is 0.055g/mL
<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
Alpha particles bouncing off of gold foil.
Answer:
..........................
Explanation:
........................
The sample of smoke described above can be described as a heterogeneous mixture. This type of mixture do not have uniform properties and composition. So, getting a certain small sample would not represent the whole mixture since it does not have uniform composition.
