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irakobra [83]
3 years ago
7

Study the scenario. The particles in a system are moving around very slowly. A few minutes later, the particles are moving, on a

verage, much faster. How does this change in motion affect the temperature of the system? the temperature increase since the average kinetic energy of the particles increases the temperature remains the same since temperature is a measure of potential energy of a system, not the movement of its particles it is not possible to tell since temperature is a result of the fastest moving particles in a system, not the average the temperature decreases since the average kinetic energy of the particles increases
Chemistry
1 answer:
Bogdan [553]3 years ago
4 0

Answer:

The average kinetic energy of the system has increased as a result of the temperature increasing.

Explanation:

Assuming this is a gas based on the framing.

The molecules of a gas span a distribution of speeds, and the average kinetic energy of the molecules is directly proportional to the absolute temperature of the sample. KEavg is proportional to T.

This can be further studied until the Kinetic-Molecular Theory.

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If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting soluti
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Answer : The boiling point of the resulting solution is, 100.6^oC

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

or,

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^oC

k_b = boiling point constant  = 0.52^oC/m

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

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w_1 = mass of solvent (water) = 10.0 g

M_2 = molar mass of solute (sucrose) = 342.3 g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}

T_b=100.6^oC

Therefore, the boiling point of the resulting solution is, 100.6^oC

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Explanation:

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<h3>Hydrogen bond</h3>

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Find out more on Hydrogen bond at: brainly.com/question/12798212

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2 years ago
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