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Dovator [93]
2 years ago
12

Be sure to answer all parts. Predict the effect of decreasing the temperature on the amounts of reactants in the following react

ions: (a) C2H2(g) + H2O(g) ⇌ CH3CHO(g) ΔH o rxn = −151 kJ The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (b) CH3CH2OH(l) + O2 (g) ⇌ CH3CO2H(l) + H2O(g) ΔH o rxn = −451 kJ The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (c) 2 C2H4(g) + O2(g) ⇌ 2 CH3CHO(g) (exothermic) The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (d) N2O4(g) ⇌ 2 NO2(g) (endothermic) The amounts of reactants decrease. The amounts of reactants stay the same. The amounts of reactants increase.
Chemistry
1 answer:
lianna [129]2 years ago
3 0

Answer:

(a) The amounts of reactants decrease.

(b) The amounts of reactants decrease.

(c) The amounts of reactants decrease.

(d) the amounts of reactants increase.

Explanation:

  • Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

<em>(a) C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) ΔH°rxn = − 151 kJ.</em>

  • Since, ΔH is a negative value, the reaction is exothermic reaction.

So, the reaction can be represented as:

C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

<em>(b) CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) ΔH°rxn = − 451 kJ.</em>

  • Since, ΔH is a negative value, the reaction is exothermic reaction.

So, the reaction can be represented as:

CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amount of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

<em>(c) 2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) (exothermic).</em>

  • Since, the reaction is exothermic reaction.

So, the reaction can be represented as:

2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

<em>(d) N₂O₄(g) ⇌ 2NO₂(g) (endothermic).</em>

  • For endothermic reaction, it can be represented as:

N₂O₄(g) + heat ⇌ 2NO₂(g)

As the temperature is decreased, it is like that decreasing the concentration of reactants side, that shifts the reaction towards the lift side (reactants side) to attain the equilibrium again. So, the amounts of reactants increase.

So, the right choice is reaction is: the amounts of reactants increase.

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
ira [324]

Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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2 years ago
Robyn has been working on developing a new task to test the reaction time of athletes. As she begins to test athletes using her
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Answer:

Ceiling effect

Explanation:

A ceiling effect takes place when the test items are not challenging enough for a group of individuals. Thus, the test score will not increase for a subsample of people who may have improved because they have already attained the highest score that can be achieved on that test.

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Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

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2 years ago
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Explanation:

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