Given what we know, the tool in question that will help the student collect data regarding the transfer of kinetic energy between water and ice would be a thermometer.

<h3>How does the thermometer measure kinetic energy?</h3>

It does not do so directly. However, kinetic energy in water molecules is reflected in the temperature of the water. When water molecules increase their kinetic energy and move more, they become hotter. Increased or decreased heat is an indirect way to measure the transfer of kinetic energy in water.

Therefore, given that the temperature of the water is a reflection of the transfer of kinetic energy happening, we can confirm that the tool that will help the student collect the data needed is a thermometer.

To learn more about kinetic energy visit:

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3 0

3 years ago

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0

3 years ago

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol C8H18

AnnyKZ [126]

<span>Enthalpy change is the total heat that is absorbed or evolved in a certain reaction that is being carried in a constant pressure system. It can be calculated by the product of the specific heat of the substance and the difference of the final temperature and the initial temperature. For this case, the enthalpy of combustion is given, so we simply multiply the amount of octane present to the enthalpy given to determine the total heat that is released.

Heat = 75 g ( 1 mol / 114.23 g ) (- 5500 kJ / mol ) = - 3611.14 kJ

The negative sign in the calculated value represents that this heat is being released in the process.</span>

6 0

3 years ago

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What would you predict for the lengths of the bonds in no2− relative to n−o single bonds and double bonds?

djverab [1.8K]

The bonds in no2− relative to n−o

8 0

3 years ago

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