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2 years ago

How many atoms are there in each of the following

1 answer:

4 0

<h3>Answer:</h3><h3>a) 9.033 × 10²³ particles</h3><h3>b) 4.068 × 10²⁴ particles</h3><h3>c) 1.51 × 10²³ particles</h3>

Explanation:

For us to answer these questions, we have to know two formulas:

Number of particles = moles × Avogadro's Number
Moles = Mass ÷ Molar Mass

Therefore:

a) particles of Na = 1.50 mol × (6.022 × 10²³) particles/mol

= 9.033 × 10²³ particles

b) particles of Pb = 6.755 mol × (6.022 × 10²³) particles/mol

= 4.068 × 10²⁴ particles

c) particles of Si

= (7.02 g ÷ 28.085 g/mol) × (6.022 × 10²³) particles/mol

= 1.51 × 10²³ particles

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

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The change in entropy is -1083.112 joules per kilogram-Kelvin.

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If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: