Answer : The solubility of this compound in g/L is 
.
Solution : Given,
Molar mass of 
= 114.945g/mole
The balanced equilibrium reaction is,
At equilibrium s s
The expression for solubility constant is,
![K_{sp}=[Mn^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BMn%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
Now put the given values in this expression, we get
The value of 's' is the molar concentration of manganese ion and carbonate ion.
Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.
Therefore, the solubility of this compound in g/L is .
Explanation:
the air particales are always moving so it also takes the smell of the chesse with it and sreads everywhere in the house.
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
Answer:
C.) That all matter was composed of earth, fire, water and air
Explanation:
Just took the test
