Answer: 404.04 kJ.
Explanation:
To calculate the moles, we use the equation:
moles of
![\Delta H=-1036kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D-1036kJ)
According to stoichiometry :
2 moles of
on burning produces = 1036 kJ
Thus 0.78 moles of
on burning produces =
Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.
Covalent compounds generally have low boiling and melting points, and are found in all three physical states at room temperature. Covalent compounds do not conduct electricity; this is because covalent compounds do not have charged particles capable of transporting electrons.
Answer:
The NaCl concentration will be 0.03 M.
Explanation:
Given data:
Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)
Initial concentration = M₁= 0.5894 M
Final volume = V₂= 1.20 L
Final concentration = M₂= ?
Solution:
By diluting the solution volume of solution will increase while number of moles of solute remain the same.
Formula:
Initial concentration × Initial volume = Final concentration × Final volume
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
M₂ = 0.5894 M × 0.05698 L / 1.20 L
M₂ = 0.0336 M /1.20
M₂ = 0.03 M
∆H° of the following reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
<h3>What is Bond Enthalpy? </h3>
The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.
It can be calculated as:
∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.
In order to Calculate ∆Hrxn for the given equation we have:
Bond energies in kJ/mol
- H—H = 436
- H—I = 295
- I—I = 151
Now, the given reaction is
H₂(g) + I₂(g) → 2HI(g)
Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.
Therefore,
We know that,
∆Hrxn = B. E(H—H) + B. E(I—I) - 2B. E(H—I)
= 436 + 151 - 2× 295
= 436+ 151 - 590
∆Hrxn = -3kJ/mol.
Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
learn more about Bond energy:
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Triglyceride. Specifically, glucose.