For an engine whose displacement is listed as 430 in.², the engine piston displacement in liters is mathematically given as
PD= 7.37 L
<h3>What is the engine piston displacement in liters of an engine whose displacement is listed as 430 in.²?</h3>
Generally, the equation for the dimensional analysis method,\, we convert in to L is mathematically given as
l*(v/l)*l/v
Therefore, piston displacement
PD=(450 in3) . (1 dm3 / 61.024 in3) . (1 L / 1 dm3)
PD= 7.37 L
In conclusion, piston displacement
PD= 7.37 L
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Hi so from what I can see the pizza as a distant or and you just have to convert the grams of glucose into moles. Most teachers ask for this format
The answer is true the answer has to be 20 words long so true true true true true true
Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
= 
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
= 
= ![1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'](https://tex.z-dn.net/?f=1%2891300%20J.mol%5E%7B-1%7D%20%29%20%2B%5Cint%5Climits%5E%7B435%7D_%7B298.15%7D%20%5B%7B%2829.86%29-%5Cfrac%7B1%7D%7B2%7D%2829.38%29-%5Cfrac%7B1%7D%7B2%7D29.13%7D%5DJ.K%5E%7B-1%7D.mol%5E%7B-1%7D%20%5C%2C%20dT%27)
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
Answer:
Here's what I get
Explanation:
1. Names
I. CH₃-CH₂-COOH = 49. propanoic acid
II. CH₃-CH₂-OH = 46. ethanol
III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate
IV. H-O-CH₂-CH₂-CH₃ = 48. propan-1-ol
V. H-COO-CH₃ = 51. methyl methanoate
VI. CH₃-COOH = 50. ethanoic acid
2. Precursors
52. methyl propionate ⇒ methanol + propanoic acid
53. ethyl methanoate ⇒ ethanol + methanoic acid
