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3 years ago

Can y'all please help me thank you

Chemistry

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

When water moves through the water cycle, it changes between these states of matter over and over again. The water cycle is the process that moves water between the air and Earth's surface. ... With enough energy, the molecules of liquid water change into water vapor and move into the air. Think of the many ways that the hydrosphere and the atmosphere connect. ... The geosphere, in turn, reflects the sun's energy back into the atmosphere. The biosphere receives gases, heat, and sunlight (energy) from the atmosphere. It receives water from the hydrosphere and a living medium from the geosphere.

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The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan

aliina [53]

Answer:

$K^{2000K}=0.774\\\\K^{3000K}=12.56$

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

$\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}$

Thus, at 2000 K:

$\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol$

And at 3000 K:

$\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol$

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

$K=exp(-\frac{\Delta _rG}{RT} )$

Thus, at each temperature we obtain:

$K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56$

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0

3 years ago

When two or more refrigerants are mixed, what must be done with the mixture?

Lana71 [14]

When refrigerants are mixed, somebody must recover the mixture into a separate container or tank.
Refrigerant is a mixture, that can transform from liquid to gas and vice versa. Refrigerants used in a heat pump and refrigerators. They should not be mixed together, one of the reason is because they are flammable.

5 0

4 years ago

If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron

OLga [1]

Answer:

$91°C$

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 Â°c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in Â°c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

Molecular mass of water= 18 g/mol

Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

= 2.322 Kj

=2322J

We can now calculate Heat gained by Iron block

Q = mCΔT

m = mass of substance

c = specific heat capacity

=change in temperature

m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

ΔT= change in temperature

T2 = T1+ ΔT

= 68.8 + 22

= 90.8 °C

=91°C

Therefore, final temperature is $91°C$

6 0

3 years ago

A quantity of water vapor at 100°C is condensed to liquid water at the same temperature. In the process, 180,414.08 joules of

salantis [7]

Answer:

0.08kg

Explanation:

Given parameters:

Temperature of water = 100°C

Quantity of heat released = 180414.08J

Unknown:

Quantity of water condensed = ?

Solution:

This is simple phase change process without any attendant change in temperature.

To solve this problem, use the formula below;

H = mL

H is the heat released

m is mass

L is the latent heat of condensation of water = 2260 kJ/kg

Insert the parameters and solve;

180,414.08 = m x 22.6 x 10⁵ J/kg

m = $\frac{180414.08}{2260000}$ = 0.08kg

5 0

3 years ago

A mixture of helium and nitrogen gases, in a 7.03 L flask at 17 °C, contains 0.738 grams of helium and 8.98 grams of nitrogen. T

Andrei [34K]

Answer:

The partial pressure of nitrogen in the flask is 1.08 atm and the total pressure in the flask is 1.70 atm.

Explanation:

We must use the Ideal Gas Law to solve this:

Pressure . volume = n . R . T

T = T° in K → T°C + 273

17°C + 273 = 290K

n = moles

In a mixture, n is the total moles (Sum of each mol, from each gas)

Moles = Mass / Molar mass

Moles He = 0.738 g / 4g/m = 0.184 moles

Moles N₂ = 8.98 g / 28g/m = 0.320 moles

0.184 m + 0.320m = 0.504 moles

P . 7.03L = 0.504m . 0.082L.atm/ mol.K . 290K

P = (0.504m . 0.082L.atm/ mol.K . 290K) /7.03L

P = 1.70 atm - This is the total pressure.

To know the partial pressure of N₂ we can apply, the molar fraction:

Moles of N₂ / Total moles = Pressure N₂ / Total pressure

0.320m / 0.504m = Pressure N₂ / 1.70atm

(0.320m / 0.504m) . 1.70atm = Pressure N₂

1.08atm = Pressure N₂

8 0

4 years ago

Can y'all please help me thank you ​

Can y'all please help me thank you