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LUCKY_DIMON [66]
2 years ago
11

Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9

Chemistry
1 answer:
Lina20 [59]2 years ago
4 0

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

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The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.
xxMikexx [17]

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

6 0
2 years ago
A molecular biologist measures the mass of cofactor a in an average yeast cell. the mass is 41.5 pg . what is the total mass in
Ronch [10]
In this question, you are given the average cofactor mass per cell (41.5pg) and the total cells count(105 cells). You are asked how much cofactor that will be found from those cells(microgram= 10^6 picogram). Then the calculation would be:
Cofactor mass= cofactor per cell * cell count= 41.5pg/cell * 105 cells= 4357.5pg= 4.36 x 10^3pg

Then convert the picogram(pg) into microgram: 4.36 x 10^3pg/ (10^6pg/microgram)= 4.36x10^-3 microgram or 0.00436 microgram

if 105 cells mean 10^5 cells, the answer should be 4.15 microgram
6 0
2 years ago
Read 2 more answers
When 0.270 mol of a nondissociating solute is dissolved in 410.0 mL of CS2, the solution boils at 47.52 ∘C. What is the molal bo
grandymaker [24]

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the <em>boiling-point elevation formula</em>:

  • <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.

molality = mol solute / kg solvent

  • Density of CS₂ = 1.26 g/cm³
  • Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we <u>solve for Kb</u>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

  • 47.52 °C - 46.3 °C = Kb * 0.5226 m
  • Kb = 0.428 m/°C
3 0
2 years ago
5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu
kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
2 years ago
Which of the following pairs of compounds would be most easily separated by thin layer chromatography: n-octyl alcohol and 1-oct
vivado [14]

Answer:

B. n-octyl alcohol and 1-octene

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. The principle is that different compounds in the sample mixture travel at different rates due to the differences in interactions with stationary phase and due to the differences in solubility in the solvent. The principal chemical property for separation using this technique is molecular polarity

You can intuit than hexadecane and octadecane don't have big polarity differences, also chlorobenzene and bromobenzene haven't.

An alcohol as n-octyl alcohol has different polarity than an alkene as 1-octene.

Thus, using  thin layer chromatography is most easy to separate:

<em>B. n-octyl alcohol and 1-octene </em>

<em></em>

I hope it helps!

<em></em>

8 0
2 years ago
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