Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
The answer is (2) A bond is formed and energy is released. The left side of equation is I atom and the right side of equation is I2 molecule. So the bond is formed between I atom to form I2 molecule. And forming bond will release energy while breaking bond will absorb energy.
The likely thing which happens when two atoms of this element move toward each other is covalent bonding.
<h3>What is Covalent bonding?</h3>
This involves the atoms of element sharing electrons in order to achieve a stable octet configuration.
The element is oxygen which has an atomic number of 8 and needs two electrons to complete its outermost shell which results in the formation of two covalent bonds.
Read more about Covalent bonding here brainly.com/question/3447218
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Answer:
No effect.
Explanation:
Hello,
In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.
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Answer: The value of equilibrium constant for new reaction is 
Explanation:
The given chemical equation follows:
The equilibrium constant for the above equation is 
We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:
The equilibrium constant for this reaction will be the cube of the initial reaction.
If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
Hence, the value of equilibrium constant for new reaction is
