Answer:
its c ma man .
This is the other ray that will make up the angle ∠AYZ and will complete the constructioExplanation:
if your taking the quick check these were my awnsers
1.c
2.b
3.d
4.c
5 .A
for the copying an angle quick check only
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: 10
ExplanAtion: My random idiot friend just told me
Answer:
Their amount is not consistent at all parts of the atmosphere
Explanation:
Carbon dioxide and water vapor exists as variable gases in the atmosphere because their amount varies from one part to the other.
- The abundance of variable gases is influenced strongly by activities from the surface.
- This variability occurs a result of the contribution from from life activities on the surface.
- Also, due to properties of the gases, they are not evenly distributed.
- At certain altitudes, the amount of water vapor can be low.
<span>The two factors that determine whether or not a molecule is polar are if the individual bonds are even and the </span>shape<span> of the molecule. If the molecule is perfectly symmetric, the molecule will not be polar even if there are polar bonds present.</span>
