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kirill115 [55]
3 years ago
6

Compose one to two typewritten paragraphs summarizing the content of your poster. a) Explain in detail what happens as thermal e

nergy is added to the sealed containers. i. How does an increase in thermal energy affect the kinetic energy of the particles in each state of matter? ii. How does an increase in thermal energy affect the particle motion in each state of matter? iii. How does an increase in thermal energy affect the temperature of each state of matter? iv. How does an increase in thermal energy affect the state of matter? b) Explain in detail what happens as thermal energy is removed from the sealed containers. i. How does a decrease in thermal energy affect the kinetic energy of the particles in each state of matter? ii. How does a decrease in thermal energy affect the particle motion in each state of matter? iii. How does a decrease in thermal energy affect the temperature of each state of matter? iv. How does a decrease in thermal energy affect the state of matter?
Chemistry
1 answer:
vladimir2022 [97]3 years ago
8 0

sad. no one answered this

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2
nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

5 0
4 years ago
Many homes that are not
andrew11 [14]

Answer:

pump that is the answer hope its right

6 0
3 years ago
Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha
DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

CaO(s)+CO_2\rightarrow CaCO_3(s)     \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)    \Delta H_1=-812.8kJ  

(2) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)     \Delta H_2=-1269kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0
3 years ago
If you have a balloon inside a car at noon during a hot summer day, the balloon molecules
aleksandrvk [35]

Answer:

it decreases

Explanation:

because its false

6 0
3 years ago
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Finally, it is time to return to Earth and review how scientists here classify organisms:
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5. Eubacteria

6. Plantae

7. Animalia

8. Protist (technically not a kingdom)

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