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Maru [420]
3 years ago
8

ASAP I WILL GIVE BRAINLIEST!Which arrow represents the change of state described above? The diagram shows changes of state betwe

en solid, liquid, and gas. The atoms of a substance lose energy during a change of state. After the change, the atoms are close together but are able to slide past one another.
0 L
0 N
0 O
0 P​

Chemistry
1 answer:
lilavasa [31]3 years ago
7 0

Answer:

O

Explanation:

The atoms lose energy during a change of state, but can still slide past each other; gas to a liquid.

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Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

4 0
3 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =
Lady bird [3.3K]

Answer:

4.86\times10^{-7}\ \text{m}

Explanation:

R = Rydberg constant = 1.09677583\times 10^7\ \text{m}^{-1}

n_1 = Principal quantum number of an energy level = 2

n_2 = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}

The wavelength of the light emitted is 4.86\times10^{-7}\ \text{m}.

6 0
3 years ago
When measuring the volume of an irregularly shaped object you would measure the amount of water____ or pushed away
Over [174]
Displacement is the answer
3 0
3 years ago
For the following hypothetical reaction: 3A + 3B —> 2C. How many moles of B will you need to convert 4 moles of A into as man
galben [10]

Answer:

Do you still need this??

Explanation:

7 0
3 years ago
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