Answer is .47
Fourty seven hundredths in standard form is .47,
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
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Answer:
219
Step-by-step explanation:
a is the number needed to add from the first to the second which is 4(must be constant)
b is the term zero means 15-a which is 15-4=11
-5/8 ahhhhhhhhhhhhhhhhhhhhhhhhh
