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:
The correct answer is option B.
Explanation:
As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.
After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.
Answer:
A
Explanation:
pls mark brainliest right
1. L
Number one because the lines match up