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cestrela7 [59]
3 years ago
6

Can you please help me to answer this? I give you a brainlest! ​

Chemistry
1 answer:
zysi [14]3 years ago
5 0
9 is B
6 is A
i’m not sure about the rest
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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
Write the symbol of the element that has an electron configuration of 1s22s22p6.
antoniya [11.8K]
Neon is the element - Ne the symbol
8 0
3 years ago
Part a the disproportionation of mn2+(aq) to mn(s) and mno2(s) in acid solution at 25 ∘c. calculate δg∘rxn.
ollegr [7]

Answer:

ΔG^{o} = -461198.3 J

Explanation:

Gibbs free energy is defined as the energy associated with a given chemical reaction which can be used to do work. Firstly, we need to figure out the chemical equation for the given problem. For the given problem, the chemical equation is:

Mn^{2+} _{(aq)} + 2e^{-}⇒Mn_{(s)}      E^{o} = -1.18 V

Mn^{2+} _{(aq)} + H_{2}O ⇒MnO_{2(s)} +4H^{+} + 2e^{-}E^{o} = -1.21 V

The addition of the two E^{o} gives the E_{cell} of the equation, i.e. -1.18-1.21 = -2.39 V.

Then, using the equation for ΔG^{o}, we have:

ΔG^{o} = n*F*E^{o} = 2*96485*-2.39 = -461198.3 J

4 0
4 years ago
What effect does an increase in products have on the reaction rate of a mixture at equilibrium? A. The forward reaction rate inc
Anna007 [38]

<u>Answer:</u>

<em>D. The reverse reaction rate increases.</em>

<u>Explanation:</u>

The<em> rate of reverse</em> increases if the products are increased in a mixture at equilibrium.

At first,<em> rate of forward and backward</em> reactions are same for a <em>mixture in equilibrium. </em>

<em>If we add products or,the reaction moves to the opposite side I.e towards reactants side.</em>

so,according to the Le Chateliers principle,the reaction shifts opposite to the side of increase. so,according to the Le Chateliers priciple, <em>the rate of reverse or backward reaction will increase with the increase in the products.</em>

4 0
3 years ago
What evidence have you discovered to explain how a wave interacts with matter?
Y_Kistochka [10]

Answer:

Three ways that waves may interact with matter are reflection, refraction, and diffraction. Reflection occurs when waves bounce back from a surface that they cannot pass through. Refraction occurs when waves bend as they enter a new medium at an angle and start traveling at a different speed.

Explanation:

7 0
3 years ago
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