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vfiekz [6]
2 years ago
14

HELP!!!

Chemistry
2 answers:
pshichka [43]2 years ago
4 0

Answer:

2.53kpa

Explanation:

p1*v1/t1 = constant

the constant can be changed to p2*v2/t2

v2 can be removed from equation as it is constant

Kitty [74]2 years ago
3 0

Answer:

2.54kpa

Explanation:

p*v/t = constant

6.58/539=p2/2111

p2=211(6.50/539)

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Answer:

b

Explanation:

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3 years ago
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
mina [271]

Answer:

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

Explanation:

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:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

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

7 0
2 years ago
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What happens when two hydrogen atoms enter the ETS as part of either NADH or FADH2?
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For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an excepti
Maksim231197 [3]

Answer:

0,0,0,0

Explanation:

The formal charge formula:

Formal charge= Valence electrons - lone pair electrons - 0.5 shared electrons

So:

Hydrogen: 1 elec. of valence and shares two electrons with the O

Formal charge= 1 - 0 - 0.5*2=0

Oxygen: 6 elec. of valence, 2 lone pairs and shares two electrons with the H and two with the F

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Note: the dative bond between F and the second O doesn't count as shared electrons.

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