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max2010maxim [7]
3 years ago
14

The change in entropy, Δ????∘rxn , is related to the the change in the number of moles of gas molecules, Δ????gas . Determine th

e change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. K(s)+O2(g) ⟶ KO2(s)
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0

Answer:

The entropy decreases.

Explanation:

The change in entropy in a chemical reaction is lower when the moles of the gas product decrease. In the same way, the change in entropy is higher when the moles of the gas product increase.

For the reaction:

K(s) + O₂(g) ⟶ KO₂(s)

The number of moles of gas reactant is one (1 mole of K and 1 mole of O₂) and there are not moles of gas in product. The change in moles of gas is 1.

As the number of moles of product decrease <em>the entropy decreses.</em>

<em></em>

I hope it helps!

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Answer:

w = 164.62 g

Explanation:

molarity of a solution is given as -

Molarity (M)  =  ( w / m ) / V ( in L)

where ,

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From the question ,

M = 500 mM = 0.5 M

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V = vol.of solution = 1 L

w = ?

<u>To find the value of w , using the above formula , and putting the respected values , </u>

Molarity (M)  =  ( w / m ) / V ( in L)

0.5  =  ( w / 329.24 ) / 1 L

w = 164.62 g

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Frequency = 2. 41×10^15 Hz

Work function =  6. 53 eV = 6. 53 * 1.6 * 10^-19 = 10.45 * 10^-19

ENERGY = Work function +  MAX KINETIC ENERGY

K max = 6.626 * 10^-34 * 2. 41×10^15 - 10.45 * 10^-19

K max = 15.96 *10^-19 - 10.45 * 10^-19

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Hence, The minimum de Broglie wavelength of the ejected electrons = 0.66 nm

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