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2 years ago

help please an btw this isn't a test ill give brainliest science is really my thing. btw help with number 6

Chemistry

1 answer:

Dmitriy789 [7]2 years ago

8 0

Since you have an internal body temperature, which circulates warmth, anything cold that is in contact with your body will start to ‘decrease’ that internal body temp (in that particular part of the body) the hot energy is transferred to the cold -> the cold energy is transferred to the hot

ex: you grab an ice cube, the ice cube begins to melt because the warmth of your hand melts it, afterwards your hand is cold

hope this makes sense ! :-)

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Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou

Georgia [21]

We know,

$\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol$

For given reaction, $I_2(g)+Cl_2(g)\ -->\ 2ICl(g)$

$\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol$

For , 2.41 moles of $I_2$ :

$\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ$

We know :

$\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K$

Hence, this is the required solution.

7 0

3 years ago

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

3 0

3 years ago

Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o. what is the molecular formula for t

Gnesinka [82]

Assuming we have 100 g of sample

30.45/MW of N 14g = 2.175

69.55/MW of O 16g = 4.34

4.34/2.185 = 2

for every 1 mole of N we have 2 moles of O

so the empirical formula would be NO2

without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question

6 0

2 years ago

Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s

Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

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3 years ago

Which of the following best defines the internal energy of a system?

Anarel [89]

<span>In thermodynamics, the internal energy of a thermodynamic system, or a body with well-defined boundaries, denoted by U, or sometimes E, is the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energy of atoms within molecules or crystals. It includes the energy in all the chemical bonds, and the energy of the free, conduction electrons in metals.</span>

7 0

2 years ago