Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
The answer is na sodium, bromine
When the salt AgI dissolves in water it dissociates as follows;
AgI ---> Ag⁺ + I⁻
molar solubility is the number of moles that are dissolved in 1 L of solution.
If molar solubility of AgI is x, then molar solubility of Ag⁺ is x and I⁻ is x.
the formula for solubility product constant - ksp of AgI is given below
ksp = [Ag⁺][I⁻]
ksp = (x)(x)
ksp = 8.51 x 10⁻¹⁷
therefore,
x² = 8.51 x 10⁻¹⁷
x = 9.22 x 10⁻⁹
since molar solubility of AgI is x, then molar solubility of AgI is 9.22 x 10⁻⁹ M
