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butalik [34]
3 years ago
13

A 1.87 mol sample of Ar gas is confined in a 45.5 liter container at 23.6 °C.

Chemistry
1 answer:
lora16 [44]3 years ago
5 0

Answer:

a. increase

Explanation:

Based on the kinetic molecular theory of gases, the average kinetic energy of the system will increase.

  • The average kinetic energy is heat
  • If temperature increases, heat of a system will also rise.
  • According the kinetic molecular theory "the temperature of the gas is a measure of the average kinetic energy of the molecules"

Therefore, due to the increase in temperature, the average kinetic energy of the system increases.

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Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.
insens350 [35]
Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.  

Required:
Heat produced in kilojoules  

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.  
 
Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles  
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or 1.015 kilojoules</u> <span>This is the amount of heat produced in warming 125 g f iron.</span>
7 0
3 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Leokris [45]
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
6 0
3 years ago
What kingdom is Spirogyra in?
jasenka [17]
Protista .................................
8 0
3 years ago
Calculate the energy E of a sample of 3.50 mol of ideal oxygen gas (O2) molecules at a temperature of 310 K. Assume that the mol
love history [14]

Answer:

13.53 kJ

Explanation:

The energy of a gas can be calculated by the equation:

E = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.

E = (3/2)*3.5*8.314*310

E = 13,531.035 J

E = 13.53 kJ

7 0
2 years ago
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