1cc (cubic centimeter) and 1mL (milliliter) are the same volume. So, 25mL = 25cc
Answer:
The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together. ... When displacement vectors are added, the result is a resultant displacement.
Answer:
The pH of a solution is simply a measure of the concentration of hydrogen ions,
H
+
, which you'll often see referred to as hydronium cations,
H
3
O
+
.
More specifically, the pH of the solution is calculated using the negative log base
10
of the concentration of the hydronium cations.
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
−
log
(
[
H
3
O
+
]
)
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Now, we use the negative log base
10
because the concentration of hydronium cations is usually significantly smaller than
1
.
As you know, every increase in the value of a log function corresponds to one order of magnitude.
Explanation:
Answer:
<u>One lone-Pair is present in Ammonia</u>
<u></u>
Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.
The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
