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:
B. Developing theories using many lines of evidence
Workplace Hazardous Materials Information System is the answer to this question. Hope it helps :)
Answer:
11.0 dm³
Explanation:
From the question,
Applying
PV= nRT............... Equation 1
Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.
make V the subeject of the equation
V = nRT/P............. Equation 2
But,
Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')
n = m/m'....................... Equation 3
Substitute equation 3 into equation 2
V = mRT/Pm'............. Equation 4
Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g
Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.
Substitute these values into equation 4
V = (301×18.4×0.082)/(32×1.3132)
V = 454.1488/42.0224
V = 10.81 dm³
V = 11.0 dm³
