Answer:
1.23 j/g. °C
Explanation:
Given data:
Mass of metal = 35.0 g
Initial temperature = 21 °C
Final temperature = 52°C
Amount of heat absorbed = 320 cal (320 ×4.184 = 1338.88 j)
Specific heat capacity of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = 52°C - 21 °C
ΔT = 31°C
1338.88 j= 35 g ×c× 31°C
1338.88 j= 1085 g.°C ×c
1338.88 j/1085 g.°C = c
1.23 j/g. °C = c
Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The combustion of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

Now we are reversing reaction 1 and then adding both the equations, we get :
(1)

(2)

The expression for enthalpy change for the reaction will be,



Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
The atomic number is 8
Atomic number=number of proton or number of electrons
Answer:
Answer is on the foto.
Explanation:
Complete Lewis formula is on picture 2