Answer:
D. Intramolecular covalent bond
Explanation:
Compound D is structurally more rigid as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules together (intermolecular bonding). On the other hand Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative atom like oxygen, Flourine,Chlorine to hydrogen.
Answer:
divergent
Explanation:
Plates move apart at a divergent boundary
Answer:
Elements that fall between those on the left and right sides of the periodic table
Explanation:
Transition metals:
These are present at the center of periodic table.
These are d-block elements.
They include the elements of group 3 to 12 in periodic table.
They have large charge to radius ratio.
They mostly form paramagnetic compounds.
They shoes more than one oxidation state.
They form colored compounds.
They all have high melting and boiling point.
They have high densities.
They form stable complexes.
The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.
The Texas City Refinery explosion occurred on March 23, 2005, when a hydrocarbon vapor cloud was ignited and violently exploded at the ISOM isomerization process unit at BP's Texas City refinery in Texas City, Texas, killing 15 workers, injuring 180 others and severely damaging the refinery.
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³