Answer:
d = 1.8 × 10⁻⁴ kg/L
Explanation:
Given data:
Moles of helium = 2 mol
Temperature and pressure = standard
Density = ?
Solution:
PV = nRT
V = nRT/ P
V = 2 mol × 0.0821 atm. L/ mol. K × 273 K / 1 atm
V = 44.8 L
Mass of helium:
Mass = number of moles × molar mass
Mass = 2 mol × 4 g/mol
Mass = 8 g
Mass = 0.008 kg
Density;
Density = mass/ volume
d = 0.008 kg/ 44.8 L
d = 1.8 × 10⁻⁴ kg/L
Answer:
Liquid volume is usually measured using either a graduated cylinder or a buret.
Answer:
A
Explanation:
Hot air is less dense therefore it rises. Cold air is more closely packed therefore it is more dense. Less dense things rise while more dense things sink.
Answer:
The ratio of the number of atoms of gold and silver in the ornament is 197 : 1080.
Explanation:
Let the mass of silver ornament be x.
Mass of gold polished on an ornament = 1% of x= 0.1 x
Moles of silver =
Number of atoms = 
Where :
= Avogadro number
Moles of gold =
Silver atoms =
Gold atoms =
The ratio of the number of atoms of gold and silver in the ornament:

= 197 : 1080
Explanation:
Ionization equation for
is as follows.

s s s
Now, the expression for the solubility product is as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 
= 
As the concentration of
is given as 0.4 M.
So,
= 0.4 M
Putting the given values as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.4)
= 12.325 \times 10^{-5}[/tex]
Hence, the solubility of
in
is
.
Therefore, solubility of
in g/ml as follows.

= 0.0167 g/L
Thus, we can conclude that solubility of
is 0.0167 g/L.