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2 years ago

At -70 ∘c and 5.2 atm, carbon dioxide is in which phase?

2 answers:

7 0

A phase diagram is a temperature vs pressure plot from which the phase of a compound can be deduced under the given temperature and pressure conditions. The 3 equilibrium curves: Solid-Gas, Liquid-Gas and Solid-Liquid separate the compound into the three states of matter. The 3 states can however coexist at a unique point referred to as the triple point.

Based on the phase diagram for CO2 (which is readily available online), at -70 C and 5.5 atm, CO2 exist as a solid.

CaHeK987 [17]2 years ago

6 0

Converting, -70 degrees centigrade is equal to 9/5 x C+32= 158 degrees Fahrenheit and cO2 freezes at -108.5 degrees Fahrenheit so the CO2 will be in solid form at this temperature or what is commonly known as the form 'dry ice'.

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What type of bond would form between two atoms of selenium?

siniylev [52]

Answer:

the correct answer is B

hope it helps !!!

8 0

2 years ago

The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG

Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate

ΔG∘ = -12.6 KJ/mole = -12600 J/mole

ΔG = -0.1 KJ/mole = -100 J/mole

[Creatine phosphate] = 25 mM = 25*10^-3 M

[Creatine] = 17 mM = 17*10^-3 M

[ATP] =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600 = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500 = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc = 6.44*10^-3

But,

Qc = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3 = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

= 2.1896 *10^-5 M

= 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0

2 years ago

What is the freezing point of a 1.40 m aqueous solution of a nonvolatile un-ionized solute? (the freezing point depression const

Alina [70]

Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m

where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m

</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC

Now, for water freezing point = 0 oC

∴Freezing point of solution = -2.604 oC

6 0

2 years ago

In the formula X2O5, the symbol X could represent an element in Group

topjm [15]

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

6 0

2 years ago

50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.

photoshop1234 [79]

Limiting reactant : O₂

Mass of N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

$\tt =\dfrac{50}{44}=1.136$

mol O₂ :

$\tt =\dfrac{50}{32}=1.5625$

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136 1.5625

1.0416 1.5625 1.0416

0.0944 0 1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

$\tt =mol\times MW=1.0416\times 92=95.83~g$

7 0

2 years ago