Answer:
(a) So the top of the hill is 365 feet above sea level.
(b) Checkpoint 5 is 185 feet higher than checkpoint 2.
Step-by-step explanation:
<u>Solution for (a):</u>
<u />
Checkpoint 2 is -218 feet above sea level.
The top of a hill rises 583 feet above Checkpoint 2.
The altitude of the top of the hill = -218 + 583 = 365
So the top of the hill is 365 feet above sea level.
<u>Solution for (b):</u>
<u />
Checkpoint 2 is -218 feet above sea level.
Checkpoint 5 is -33 feet above sea level.
Checkpoint 5 is -33 - -218 = 185 feet higher than checkpoint 2.
(-5+8*i)+(-9+5*i) evaluates to <span>-14+13i</span>
Answer:
Can you provide more context, as the standard number line shows -3 and the opposite of -3 which is +3
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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