The standard enthalpy change for the following reaction is 576 kJ at 298 K. 2 NaI(s) 2 Na(s) + I2(s) ΔH° = 576 kJ What is the standard enthalpy change for this reaction at 298 K? Na(s) + 1/2 I2(s) NaI(s)

ΔH° = -288kJ

Explanation:

You can make algebraic opperations of the chemical reations to obtain the ΔH of a particular reaction.

From:

2NaI(s) → 2Na(s) + I2(s) ΔH° = 576 kJ

The ΔH of the inverse reaction:

2Na(s) + I2(s) → 2NaI(s) ΔH° = -576 kJ

Dividing the last reaction twice:

Na(s) + 1/2 I2(s) → NaI(s) ΔH° = -576 kJ / 2

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.