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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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8 0
3 years ago
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Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
drek231 [11]

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

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2 years ago
According to Newton’s first law of motion, when will an object at rest begin to move?
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the student records the concentration of the stock solution of H2O2 to be 0.950 M. The student proceeds to prepare the reaction
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Answer:

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Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:

25.0mL = <em>0.025L of solution</em>

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And moles of KI are:

0.0100L × 0.10M = <em>0.00100 moles of KI</em>

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Thus, molarity is:

0.00100 moles / 0.025L = <em>0.0400M of KI</em>

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A student dissolves 73 g of sodium nitrate, NaNO3, in 100 mL of water and observes a clear, colorless
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