Answer : The value of 
for this reaction is 36.18 kJ
Explanation :
First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.
As per first law of thermodynamic,
where,
= internal energy of the system
q = heat added or rejected by the system
w = work done
As we are given that:
q = 38.65 kJ
w = -2.47 kJ (system work done on surrounding)
Now put all the given values in the above expression, we get:
Therefore, the value of for this reaction is 36.18 kJ
The molarity of the solution will be 0.72 m.
The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.
The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.
Given data:
Molarity can be determined by the formula:
where, M is molarity and V is volume.
Put the value of given data in above equation.
57.3 × 0.497 m = M × 39.5 L
M = 0.72 m
Therefore, the molarity of the solution will be 0.72 m
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Answer:
4) transferred from the valence shell of one atom to the valence shell of another atom
Explanation:
Electrons are located outside of the nucleus which contains the protons and the neutrons.
For bonds to form, valence electrons located in the outermost shell electrons are involved. These are the valence electrons. These outer shell electrons can be shared or transferred between two combining atoms to form stable atoms.
In ionic bonds, the electrons are transferred from one specie to another. The atom that loses the electrons becomes positively charged and the receiving atom becomes negatively charged. This is the crux of ionic bonds.
Answer:
the brain and the spinal cord are the two main parts.
Answer is: sodium (Na) and iodine (I₂).
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First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
