Answer:
are present in solution.
Explanation:
Molarity of the solution = 0.210 M
Volume of the solution = 65.5 ml = 0.0655 L
Moles of aluminum iodide= n


n = 0.013755 moles of aluminum iodide
1 mole of aluminum iodide contains 3 moles of iodide ions:
Then 0.013755 moles of aluminum iodide will contain:
of iodide ions
Number of iodide ions in 0.041265 moles:

are present in solution.
It would be gold ...................
Answer:
See explanation
Explanation:
We define the formal charge on an atom in a molecule as the charge it carries assuming that electrons in all chemical bonds of the molecule were shared equally between atoms irrespective of the electronegativity of each atom.
The formula for calculating the formal charge on an atom in a molecule is;
Formal Charge = [number of valence electrons on neutral atom] – [(number of lone electron pairs) + (½ number of bonding electrons)] ·
The formal charge on the two nitrogen atoms in diazomethane is obtained as follows;
Middle nitrogen atom = 5 – 8/2 – 0 = +1
Last nitrogen atom = 5 – 4/2 – 4 = –1
The Lewis structure of the molecule is shown in the image attached.
Answer:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
![\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}](https://tex.z-dn.net/?f=%5Ctext%7B%5BH%24%5E%7B%2B%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20H%7D%5E%7B%2B%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D)
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
![\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pOH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20OH%7D%5E%7B-%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D)
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7
Answer:
68.3 kilo Pascal is the pressure in the container.
Explanation:
To calculate the new pressure , we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:

where,
are initial pressure and volume.
are final pressure and volume.
We are given:

Putting values in above equation, we get:


68.3 kilo Pascal is the pressure in the container.