Answer:
B) The group the atom is located
C) There needs to be 2 Na atoms and 1O atom to form Na2O
Answer:
mass of platinum = 2526.12 g
Explanation:
Given data:
Mass of water = 125 g
Initial temperature of water= 100.0°C
Initial temperature of Pt = 20.0°C
Final temperature = 235°C
Specific heat of Pt = 0.13 j/g°C
Specific heat of water = 4.184 j/g°C
Mass of platinum = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Q(w) = Q(Pt)
m.c. (T2 - T1) = m.c.
(T2 - T1)
125 g × 4.184 j/g°C × (235°C - 100.0°C) = m × 0.13 j/g°C × (235°C - 20°C)
125 g × 4.184 j/g°C × 135°C = m × 0.13 j/g°C × 215°C
70605 j = m×27.95 j/g
m = 70605 j /27.95 j/g
m = 2526.12 g
Answer:

Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

The net equation is then:

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:

Actual cathode half-equation:

Actual net reaction:
