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Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

Answers

Answer 1

Answer:

Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$ $K_p=100$

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i) $NO(g)+(1)/(2)O_2(g)\rightleftharpoons NO_2(g)$ $K_p_1=?$

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:

$K_p_1=(K_p)^{(1)/(2)}$

$K_p_1=(100)^{(1)/(2)}$

$K_p_1=10$

The value of equilibrium constants for this reaction is, 10

(ii) $NO_2(g)\rightleftharpoons NO(g)+(1)/(2)O_2(g)$ $K_p_2=?$

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:

$K_p_2=(1)/((K_p_1))$

$K_p_2=(1)/((10))$

$K_p_2=0.1$

The value of equilibrium constants for this reaction is, 0.1

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Which of the following solutions would make a good buffer system? (Check all that apply.) A. A solution that is 0.10 M NH3 and 0.10 M NH4Cl B. A solution that is 0.10 M HCN and 0.10 M NaF C. A solution that is 0.10 M HCN and 0.10 M LiCN D. A solution that is 0.10 M HF and 0.10 M NaF

The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

(II) To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.872cm in diameter is to be placed in a hole 1.870cm in diameter in a metal at 22°C. To what temperature must the rivet be cooled if it is to fit in the hole

Answers

Given:

Rivet diameter, $d_(r)$ = 1.872 cm

Hole diameter, $d_(h)$ = 1.870 cm

Temperature, $T_(2)$ = 22 °C

Formula Used:

$\alpha = (\Delta d)/(d* \Delta T)$

where,

$\alpha$ = coefficient of linear expansion

$\Delta d$ = change in diameter = $d_(h) - d_(r)$

$\Delta T}$ = change in temperature = $T_(2) - T_(1)$

Solution:

we know that coefficient of linear expansion of steel, $\alpha$ = $12* 10^(-6)/^(\circ)C$

Using the above formula :

$\alpha = (\Delta d)/(d* \Delta T)$

$12* 10^(-6)/^(\circ)C$ = \frac{1.870 - 1.872}{1.872\times \T_{2} - T_{1}}[/tex]

$T_(2) - 20/^(\circ)C$ = \frac{1.870 - 1.872}{12\times 10^{-6}}}[/tex]

$T_(2) = -67.03/^(\circ)C$

Therefore, the rivet must be cooled to $-67.03/^(\circ)C$

Final answer:

The question involves the concept of thermal expansion in Physics. By knowing the initial diameter of the rivet and hole, as well as the ambient temperature, we can use the thermal expansion formula to calculate the temperature to which the steel rivet must be cooled to fit into the hole.

Explanation:

The subject in question pertains to Physics and specifically to the concept of thermal expansion. This indicates how objects (in this case, a steel rivet) tend to change in volume or shape as a response to a change in temperature. The diameter of the rivet when cooled will decrease slightly, allowing it to fit into the smaller hole.

To find the temperature to which the rivet needs to be cooled, we require knowledge of the thermal expansion coefficient of steel, which (for generalization) can be averaged to around 0.000012 (1/°C). The formula to calculate the change in diameter (Δd) is:

Δd = α * d * ΔT

where α is the coefficient of linear expansion, d is the original diameter, and ΔT is the change in temperature. Knowing the initial diameter of the rivet and the hole it must fit into, together with the ambient temperature (22°C), we can rearrange this formula to find the cooling temperature needed for the rivet to fit into the hole.

Learn more about Thermal Expansion here:

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Which of the following equations violates the law of conservation of mass? A. FeCl3 + 3NaOH yields Fe(OH)3 + 3NaCl

B. CS2 + 3O2 yields CO2 + 2SO2

C. Mg(ClO3)2 yields MgCl2 + 2O2

D. Zn + H2SO4 yields H2 + ZnSO4

Answers

Hi!

The chemical equation that violates the law of conservation of mass is C

Mg(ClO₃)₂ → MgCl₂ + 2O₂

Let's see why:

The number of atoms from each element should be the same on both sides of the equation.

For Mg:1 atoms Left Side=1 atoms Right Side
For Cl:2 atoms Left Side=2 atoms Right Side
For O: 6 atoms Left Side ≠ 4 atoms Right Side

So, the equation is not balanced for O. The correct balanced equation is the following:

Mg(ClO₃)₂ → MgCl₂ + 3O₂

Have a nice day!

What is the pOH of a solution with a [OH^ - ] of 10^ -11?

Answers

Answer:

pOH= 11

Explanation:

pOH= -log[10^ -11]= 11

The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

Answers

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL * (0.996g)/(1mL) =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O * (550gGlucose)/(398gH_2O) = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.

Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.

To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.

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Why does the mixture of Fe and S have different reactions than the compound FeS?

Answers

Answer:

fe+s is a mixture of iron and sulphur and it can be separated by magnetic separation. But FeS is a compound formed by heating iron and sulphur together.it can't be separated by any method.

Explanation:

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