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KatRina [158]
2 years ago
13

A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1°C for four-phase equilibrium of allotropic solid

forms of the exotic chemical β-miasmone. Is the claim valid or not?
Chemistry
1 answer:
Inessa [10]2 years ago
8 0

Answer:

The claim is untrue.  

Explanation:

Given the information from the question. We need to evaluate the claim.

According to the phase rule we have F= C-P+2

In this particular situation, the forms are completely allotropic .In order words, they are conjured through the same chemical composition. Thus constituents C= 1, P=4 for four phases and the number variables is 2. As a result, F= C-P+2 =1-4+2= -1. Therefore, the claim is untrue.  

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In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 27.0 ∘C . You identified that the ideal pressure
dimaraw [331]

Answer : The percent difference between the ideal and real gas is, 4.06 %.

Explanation : Given,

Ideal pressure (true value) = 49.3 atm

Real pressure (measured value) = 47.3 atm

The formula used to calculate percent difference is :

Percent difference = \frac{\text{True value - Measured value}}{\text{True value}} \times 100

Percent difference = \frac{(49.3- 47.3)atm}{49.3atm}\times 100

Percent difference = 4.06 %

Therefore, the percent difference between the ideal and real gas is, 4.06 %.

4 0
2 years ago
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
Read 2 more answers
What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?
Nonamiya [84]
<span>Answer: It depends on what came after "0.5440 M H...". If it was a monoprotic acid, like HCl, the calculation would go like this: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH If it was a diprotic acid, like H2SO4, like this: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH If it was a triprotic acid, like H3PO4, like this: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH</span>
5 0
2 years ago
Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of
seraphim [82]

Answer:

The temperature difference of the body after 3 hours = 5.16 K

Explanation:

we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹

                                   or, 1.2 mole.h⁻¹

The average heat evolved by the oxidation of foodstuffs is then:

⇒          Q avg =\frac{1.2 X 420 X 10^{3} }{70} = 7.2 kj.h⁻¹.Kg⁻¹

the heat produced after 3 h would be:

                 =    7.2 kj. h⁻¹.Kg⁻¹ x 3 h

                 = 21.6 kj. kg⁻¹

                 = 21.6 x 10³ j kg⁻¹

We know Qp = Cp x ΔT

Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹

⇒ ΔT = \frac{Qp}{Cp}

⇒ ΔT = \frac{(21.6 X 10^{3} j.kg^{-1} ) }{(4.18 j k^{-1}g^{-1})   X (1000g.kg^{-1} )}

⇒ ΔT = 5.16 K

6 0
2 years ago
By mistake, you added salt instead of sugar to the<br> oil. How can you remove the salt?
Jet001 [13]

Answer:

by adding water into the mix

Explanation:

this will dissolve the salt

5 0
2 years ago
Read 2 more answers
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