In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =
= constant
=
==> P1 * V1 = P2 * V2
V2 =
=
V2 = 0.00581 m3 = 5.81 L
From the molarity and volume of HClO4, we can determine how many moles of H+ we initially have:
0.18 M HClO4 * 0.100 L HClO4 = 0.018 moles H+
We can determine how many moles of OH- we have from the molarity and volume of LiOH:
0.27 M LiOH * 0.030 L LiOH = 0.0081 moles OH-
When the HClO4 and LiOH neutralize each other, the remaining will be
0.018 moles H+ - 0.0081 moles OH- = 0.0099 moles of excess H+
This means that the molarity [H+] will be
[H+] = 0.0099 moles H+ / (0.100 L + 0.030 L) = 0.07615 M
The pH of the solution will therefore be
pH = -log [H+] = -log 0.07615 = 1.12
Lipids cannot be compressed since there is only a small distance between the molecules when bonded
Missing question:
<span>A. [PdZn(H2O)2(CO)2]Br4.
B. [Zn(H2O)2(CO)2]2[PdBr4].
C. [Pd(H2O)2][Zn(CO)2]Br4.
D. [Pd(H2O)2]2[Zn(CO)2]3Br4.
E. [Zn(H2O)2(CO)2][PdBr4].
</span>Answer is: E. [Zn(H2O)2(CO)2][PdBr4]..
In this complex diaqua means two waters (H₂O), <span>dicarbonyl means two carbonyl groups (CO), zinc(Zn) and palladium (Pd) are central atoms or metals, bromine has negative charge -1. Bromine, water and carbonyl are ligands.</span>
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
