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3 years ago

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Consider the following system at equilibrium at 723 K: 2 NH3 (g) 26.6 kcal N2 (g) 3 H2 (g) Indicate whether each individual chan

ge would favor the production of N2 (g). Evaluate each change separately, assuming that all other conditions remain constant.

1 answer:

ZanzabumX [31]3 years ago

4 0

Answer:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

- To add more reactant in order to increase its equilibrium concentration.

Explanation:

Hello,

The undergoing chemical reaction is:

$2NH_3(g)\rightleftharpoons 3H_2(g)+N_2(g)$

Thus, in order to intensify the amount of nitrogen as the chemical reaction is endothermic, considering the Le Chatelier's principle we state:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

- To add more reactant in order to increase its equilibrium concentration.

Best regards.

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The mass of water in a single popcorn kernel was found to be 0.905 grams after it popped at a temperature of 175 °C. Using the i

Rus_ich [418]

Answer:

0.583 kilojoules

Explanation:

The amount of heat required to pop a single kernel can be calculated using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of water (g)

c = specific heat capacity of water (4.184 J/g°C)

∆T = change in temperature

From the given information, m = 0.905 g, initial temperature (room temperature) = 21°C , final temperature = 175°C, Q = ?

Q = m × c × ∆T

Q = 0.905 × 4.184 × (175°C - 21°C)

Q = 3.786 × 154

Q = 583.044 Joules

In kilojoules i.e. we divide by 1000, the amount of heat is:

= 583.04/1000

= 0.583 kilojoules

8 0

3 years ago

How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

$\\ \Large\sf\longmapsto 250\times 15\%$

$\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}$

$\\ \Large\sf\longmapsto 37.5g$

<u>37.5g of NaCl present in 250g of solution.</u>

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3 years ago

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Choose all of the options that demonstrate how humans are impacting the water cycle.

Fantom [35]

Answer:

A.) Air pollutants wash onto land and water bodies from precipitation

C.) Storm water runoff carries pollution

I hope I helped! ^-^

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2 years ago

Methane burns in the presence of oxygen to form carbon dioxide and water.

wlad13 [49]

Answer:

= 9.28 g CO₂

Explanation:

First write a balanced equation:

CH₄ + 2O₂ -> 2H₂O + CO₂

Convert the information to moles

7.50g CH₄ = 0.46875 mol CH₄

13.5g O₂ = 0.421875 mol O₂

Theoretical molar ratio CH₄:O₂ -> 1:2

Actual ratio is 0.46875 : 0.421875 ≈ 1:1

If all CH₄ is used up, there would need to be more O₂

So O₂ is the limiting reactant and we use this in our equation

Use molar ratio to find moles of CO₂

0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂

Then convert to grams

0.2109375 mol CO₂ = 9.28114 g CO₂

round to 3 sig figs

= 9.28 g CO₂

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3 years ago

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

3 0

3 years ago