there is more air near sea level
One mole is always the same number: 6.02 * 10^ 23.
So, one mole of cars = 6.02 * 10 ^23 cars; one mole of pencils = 6.02 * 10^23 pencils; one mole of atoms = 6.02 * 10^23 atoms; one mole of molecules = 6.02 * 10^23 molecules.
So, all the options are correct: one mole of calcium ions has 6.02 * 10^23 representative particles, such as one mole of calcium nuclei and one of calcium atoms.
Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation:
Answer:
The volume is 19.7 mL
Explanation:
<u>Step 1</u>: Given data
Pressure at sea level = 1.00 atm
Pressure at 50 ft = 2.47535 atm
kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm
Molarity (M) = kH x P
<u>Step 2</u>: Calculate molarity
M at sea level:
M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L
M at 50ft:
M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L
We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.
13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L
Step 3: Calculate volume
P*V=nRT
with P = 1.00 atm
with V = TO BE DETERMINED
with n = 8.04*10^-4 mol/L *1L = 8.04*10^-4
with R= 0.0821 atm * L/ mol *K
with T = 25 °C = 273+25 = 298 Kelvin
To find the volume, we re-organize the formula to: V=nRT/P
V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml
The volume is 19.7 mL
