The minimum quantity of energy that the reacting species must possess in order to undergo a specified reaction.
Answer:
Explanation:
A sound knowledge of specific heat capacity of the metals is required in this case.
The specific heat capacity of a metal is the quantity of heat required to the raise the temperature of a unit mass of it by 1°C.
It is related to quantity of heat using the expression below;
H = m c Δt
where m is the mass
c is the specific heat capacity
Δt is the temperature change
let us make the specific the subject of the expression;
c = 
we can see that there is an inverse relationship between specific heat and temperature change.
The specific heat capacity of a body is an intensive property that is unique to the metal.
The higher the specific heat capacity, the lower the amount of temperature change in it.
Let us find the specific heat capacity of the given metals;
Aluminium 0.897J/gK
Iron 0.412J/gK
Silver 0.24J/gK
After the heat is supplied,
Silver > Iron > Aluminium in terms of temperature change
Answer: In CaF2, the oxidation number of Ca is +2
, and that of F is -1
. In H2SO4, the oxidation number of H is +1
, that of S is +6
, and that of O is -2
. In CaSO4, the oxidation number of Ca is +2
, that of S is +6
, and that of O is -2
. In HF, the oxidation number of H is +1
, and that of F is -1
Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Answer:
142240
Explanation:
We are told in the question:
Height of Gateway Arch in St. Louis, MO = 630ft
We are asked, how many U.S. dimes would be in a stack of the same
height when 1 dime is 1.35 mm thick.
Step 1
Convert height in ft to mm
1 ft = 304.8 mm
630ft =
Cross Multiply
630ft × 304.8mm/1ft
= 192024 mm.
Step 2
To find how many US dimes would be in a stack of the same height
= Total thickness/ Thickness of 1 US dime
= 192024 mm/1.35mm
= 142240
Therefore, the number of dimes that would be in a stack of the same
height is 142240
