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>
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- 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>
The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required
Answer:
100 C or 212 F
Explanation:
That is the boiling point for liquid so both gas and liquid are presented.
Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.
Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9
