Answer: There are 0.00269 moles of acetic acid in buffer.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of acetic acid solution = 0.0880 M
Volume of solution = 30.6 mL
Putting values in equation 1, we get:
Thus there are 0.00269 moles of acetic acid in buffer.
Answer:
Its phosphorus (P)
Explanation:
In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
Potassium (K) is in Group I of the periodic table, and elements in the same column (period) are similar. Sodium (Na), or lithium (Li) are similar.
2
C
4
H
10
(
g
) + 13
O
2
(
g
) = 8
C
O
2
(
g
) + 10
H
2
O
(
g
)
The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:
PbI2 = Pb^2+ + 2I-
We solve as follows:
Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
</span><span>1.4 x 10-8 = x(2x)^2
</span><span>1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M
The molar solubility would be </span>1.5x10^-3 M.
