Answer:
0.186M
Explanation:
First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:
<em>Moles Nitric acid:</em>
0.0200L * (0.100mol / L) = 0.00200 moles
0.120L * (0.200mol / L)= 0.02400 moles
Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid
Molarity: 0.026 moles / 0.140L
<h3>0.186M</h3>
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.
Answer: 2-Benzyloxyethanol is produced reaction in the fig. below.
Explanation:
Answer: YES
Explanation:
POLAR BONDS will form.
A polar bond is a covalent bond between two atoms where the electrons forming the bond are unequally distributed. This causes the molecule to have a slight electrical dipole moment where one end is slightly positive and the other is slightly negative.
<h3>
Answer</h3>
HF
<h3>Explanation</h3>
A buffer solution contains <em>a weak acid</em> and<em> its conjugate base</em>. The two species shall have a similar concentration in the solution. It's also possible for <em>a weak base</em> and <em>its conjugate acid</em> to form a buffer solution.
The KF solution already contains large number of 
ions. The objective is to thus find its conjugate acid or base.
contains no proton 
and is unlikely to be a conjugate acid. Assuming that is a conjugate base. Adding one proton to would produce its conjugate acid.
Therefore 
is the conjugate acid of . happens to be a weak acid. As a result, combining with would produce a solution with large number of both the weak acid and its conjugate base, which is a buffer solution by definition.
