Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at 298.15 K. Molar heat capacities of NO(g), N2(g), and O2(g) at 298.15 K are 29.86, 29.13, and 29.38 J⋅K−1⋅mol−1. The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

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.

∴

=

=

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

≅ 91383 J

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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