Answer:
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Step-by-step explanation:
The enthalpy change (ΔH) for the reaction given the data from the question is –900.8 KJ
<h3>Data obtained from the question</h3>
- 4NH₃ + 5O₂ —> 6H₂O + 4NO
- Enthalpy of ammonia, NH₃ = –46.2 KJ
- Enthalpy of Oxygen = 0 KJ
- Enthalpy of water, H₂O = –241.8 KJ
- Enthalpy of nitric oxide, NO = 91.3 KJ
- Enthalpy change (ΔH) =?
<h3>How to determine the enthalpy change</h3>
ΔHrxn = ∑ΔH(products) - ∑ΔH(reactants)
ΔHrxn = ∑[H(H₂O) + H(NO)] - ∑[H(NH₃) + H(O₂)]
ΔHrxn = [(6 × –241.8) + (4 × 91.3)] – [(4 × –46.2) + (5×0)]
ΔHrxn = –1085.6 + 184.8
ΔHrxn = –900.8 KJ
Learn more about enthalpy change:
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Answer:
It’s true
Step-by-step explanation:
If you have two even numbers, both will have 2 as a factor. So 2 would be a factor of the GCF. With 2 a factor of the GCF, no matter what anything else is, the GCF will be even.
2/3. 1/2 is equal to 3/6 and 3/6 + 1/6 is 4/6 and 4/6 simplified is 2/3.
Answer:
A number line with a closed circle on 6 and shading to the right
Step-by-step explanation:
