As a base is added to an acidic solution, the H+ ions in solution that make it acidic are slowly neutralized into water (via OH-, the base). As these ions are converted into water the concentration of them decreases, so the pH decreases, as they are directly related.
Hope this helps!
The volume of SO2 produced at 325k is calculated as below
calculate the moles of SO2 produced which is calculated as follows
write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2
find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles
by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2
use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm
V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
<u>Answer:</u> The correct answer is 
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.
Chromium will undergo oxidation reaction and will get oxidized.
The half reactions for the above cell is:
Oxidation half reaction: 
Reduction half reaction: 
( × 3)
Net equation: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
Hence, the correct answer is
