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Natasha_Volkova [10]
4 years ago
6

The phase that occurs beyond the critical point is:

Chemistry
1 answer:
Tcecarenko [31]4 years ago
5 0

Answer:

C.  Gas

Explanation:

A supercritical fluid occurs beyond the temperature of  critical point, wherein the state  matter transitions from liquid to gaseous phase interchangeably. Whereas, triple point occurs when all three states of matter: solid, liquid nd gas; coexist.

Hence the answer to the question is Gaseous phase/

I hope you found this helpful.

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Which of the following best describes how forces affect matter?
OlgaM077 [116]

Answer:

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3 0
3 years ago
4. a compound called pyrene has the empirical formula c8h5. when 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boilin
e-lub [12.9K]

The molecular mass of pyrene is 204.4 g/mol.

From;

ΔT = Kb m i

Where;

  • ΔT = boiling point elevation
  • Kb = boiling point constant
  • m = molality
  • i = Van't Hoff factor

Since the compound is molecular; i = 1

The number of moles of pyrene = 4.04 g/MM

Where; MM = molar mass of pyrene

molality = number of moles of pyrene/mass of solvent in Kg

The mass of solvent = 10 g or 0.01 Kg

molality =  4.04 g/MM/0.01

ΔT = Boiling point of solution - Boiling point of pure solvent

ΔT = 85.1°C - 80.1°C

ΔT = 5°C

5 = 2.53 × 4.04 g/MM/0.01 × 1

5 = 10.22 × 1/0.01 MM

0.05MM = 10.22

MM= 10.22/0.05

MM= 204.4 g/mol

Learn more: brainly.com/question/2292439

3 0
2 years ago
A sample of rainwater is tested with the indicator bromothymol blue and it turns yellow. What is most likely PH of rainwater? 4,
kirill [66]

Answer:

4

Explanation:

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7 0
4 years ago
What is the name of the state when electrons absorb energy and move to a higher energy level?
Sergeu [11.5K]

Answer:

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Explanation:

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4 0
3 years ago
Read 2 more answers
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
4 years ago
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