Helllllp me please Part D ends with neon Express your answer as an integer. 20

_________________ states that when a stress is applied, a reversible reaction will undergo a shift in order to re-establish its

pochemuha

LaChatelier's Principle

LaChatelier's Principle is a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

5 0

3 years ago

Why is a very small amount of cerium oxide nanoparticles need: A) the nanoparticles are elements

irina [24]

Have a high suface area to volume ratio

7 0

3 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

3 years ago

A sample of gas occupies 17 ml at â112Â°c. what volume does the sample occupy at 70Â°c?

atroni [7]

Answer is: volume of the sample is 36,2 mL.
V₁(sample) = 17 mL.
T₁(sample) = −112°C = -112 + 273,15 = 161,15 K.
T₂(sample) = 70°C = 70 + 273,15 = 343,15 K.
V₂(sample) = ?
Charles' Law: V₁/T₁ = V₂/T₂.
17 mL/161,15 K = V₂/343,15 K.
161,15 K · V₂ = 17 mL · 343,15 K
V₂ = 5833,55 K·mL ÷ 161,15 K
V₂ = 36,2 mL.

5 0

3 years ago

Your chemistry teacher empties her dehumidifier once a day if The composite of the tank is 2 gallons determine how many molecule

Natasha_Volkova [10]

Given data:

Amount of water collected in the dehumidifier= 2 gallons

Time taken to collect 2 gallons = 1 day

Calculation

Step1 : Find the mass of 2 gallons of water

1 gallon = 3.79 L

Therefore, 2 gallons = 2 gallon * 3.79 L/1 gallon = 7.58 L

Density of water = 1 kg/L

Therefore, Mass of 7.58 L water = Density * Vol = 1 * 7.58 = 7.58 kg = 7580 g

Step 2: Convert mass of water to molecules

1 mole of water = 18 g = 6.023 *10^23 molecules

therefore, 7580 g contains: 6.023 * 10^23 * 7580/18 = 2.536 *10^26 molecules

Step 3: Calculate the molecules of water collected/millisecond

Now,

2.536*10^26 water molecules are collected in 1 day

1 day = 24 hrs * 60 min * 60 sec = 86400 sec

10^-3 sec = 1 millisec

therefore, 86400 sec = 1 millisec * 86400 sec/10^-3 sec = 8.64 * 10^7 millisec

Thus water molecules collected/millisec = 2.536*10^26/8.64*10^7

= 2.935 *10^32 molecules/millisec

6 0

3 years ago

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