That would be "conductivity". All substances have some measurable amount it, but metals have by far the most.
Answer:
0.020 kg=20 g this the largest
20.0 mg=.020g
.200g
2000ug
Explanation:
Answer:
l=1
Explanation:
The angular momentum quantum number is a quantum number that describes the 'shape' of an orbital and tells us which subshells are present in the principal shell. A critical look at the electronic configuration of phosphorus at the ground state depicts the outermost electron to be in the P orbital where the angular quantum number ,l=1 .
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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>
