This may help you
Allright for <span><span>H2</span>O:</span>
- The central atom is? --> the oxygen atom
- How many atoms are bonded to the central atom? --> 2 hydrogen atoms
- How many lone pairs of electrons are on the central atom? --> O has 6 electrons and has 2 single bonds, so 2 pairs
- How many single bonds are there in this molecule? --> 2
- How many multiple bonds (double and/or triple) are there in this molecule? --> none
For each of your molecules, answer the following questions:
1. Determine the electronegativity between the atoms of each molecule.
Electronegativity O = 3.44
Electronegativity H = 2.20
3.44-2.20=1.24, so the electronegativity between O and H = 1.24
2. Identify the bond as either ionic or covalent.
Electronegativity of 0.0-1.7 = covalent
Electronegativity of 1.7-3.3 = ionic
So it's a covalent bond
3. State whether the molecule is polar or non polar.
Electronegativity of 0.5-1.7= polar covalent
4. Identify the structure as having hydrogen bonding, dipole-dipole moments or London dispersion forces (LDF).
<span><span>H2</span>O</span><span> = hydrogen bonding</span>
Answer:
1. Decrease the pH
2. The pH remains unchanged
3. Decrease the pH
Explanation:
We can calculate the pH of a weak acid using the following expression:
[1]
where,
Ka is the acid dissociation constant
Ca is the initial concentration of the acid
<em>State whether each of the following would increase, decrease, or leave the initial pH unchanged and explain your reasoning</em>
- <em>Using the same weak acid, but having a higher concentration. </em>According to eq. [1], a higher Ca leads to a higher [H⁺] and a lower pH.
- <em>Using 80.00 mL of this weak acid instead of 40.00 mL.</em> pH refers to a concentration, which is an intensive property, so it does not change when we change the amount of matter.
- <em>Using a different weak acid that has a larger Ka value.</em> According to eq. [1], a higher Ka leads to a higher [H⁺] and a lower pH.
Answer:
A piece of DNA from the father carried the trait for blue eye color
What exactly is this asking