Answer:
- 1273.02 kJ.
Explanation:
This problem can be solved using Hess's Law.
Hess's Law states that <em>regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.</em>
- We should modify the given 3 equations to obtain the proposed reaction:
<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>
<em></em>
- We should multiply the first equation by (6) and also multiply its ΔH by (6):
6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,
- Also, we should multiply the second equation and its ΔH by (6):
6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.
- Finally, we should reverse the first equation and multiply its ΔH by (- 1):
6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.
- By summing the three equations, we cam get the proposed reaction:
<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>
<em></em>
- And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:
<em>∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ =</em> (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = <em>- 1273.02 kJ.</em>
Answer:
3 M
Explanation:
Given data
- Moles of sodium hydroxide (solute): 5 moles
- Volume of solution: 2 liters
We can calculate the molar concentration, or molarity (M), of the solution using the following expression.
M = moles of solute / volume of solution (in liters)
M = 5 moles / 2 L
M = 2.5 M ≈ 3 M (if we round off to 1 significant figure)
Answer: The answer is 4.5e+27 grams, I hope I helped!
Explanation: N/A (I threw away my paper ^^")
<span>Cations mix with anions, so you know NH4+ won't mix with K+ and SO4(2-) won't miix with F-. For the reason that NH4+ and F- together have single charges, they'll mix in a 1:1 ratio, NH4F. There's two charges on SO4(2-), so it'll need two K+ to mix with, K2SO4.</span>
The animals have to adapt to their new environment meant and continue mating so they do not go extinct
