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IgorC [24]
3 years ago
14

1. The actual yield of a product in a reaction was measured as 1.20 g. If the theoretical yield of the product for the reaction

is 1.82 g, what is the percentage yield of the product?
65.9%

67.2%

71.9%

73.3%

_______________

2. Which of the following statements best defines the percentage yield of a reaction? (5 points)


The ratio of measured yield over actual yield

The amount of product measured after a reaction

The amount of measured yield over calculated yield

The maximum amount of product that can be obtained

________________________

3. A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown.


Na + Cl2 → NaCl


If the percentage yield of the reaction is 88%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.
__________________
4. Two gas jars are connected to each other, but they are separated by a closed valve. One gas jar contains oxygen, and the other contains hydrogen. What will happen when the valve is opened?

Oxygen and hydrogen molecules will mix because of random movement of their molecules.

Oxygen and hydrogen molecules will remain in separate jars because of different partial pressures.

Oxygen and hydrogen molecules will remain in separate jars because of different rates of expansion.

Oxygen and hydrogen molecules will mix because the molecules have a much lower volume than the volume of the jars.
Chemistry
1 answer:
n200080 [17]3 years ago
5 0
<h3>1. <u>Answer;</u></h3>

Percentage yield = 65.9%

<h3><u>Explanation;</u></h3>
  • Percentage yield of a reaction is obtained by dividing the actual yield by theoretical yield and multiplying by 100 percent.

Therefore;

  • Percentage yield = Actual yield/theoretical yield × 100%

                                      = (1.20 g/1.82 g) × 100%

                                      =  65.9%

<h3 /><h3>2. <u>Answer;</u></h3>

The amount of measured yield over calculated yield

<h3>Explanation;</h3>
  • Percentage yield of a reaction is obtained by dividing the actual yield by theoretical yield and multiplying by 100 percent.
  • The actual yield is the actual amount produced when the experiment or reaction is carried out.
  • Theoretical yield is the calculated or expected amount of the product. It is based on the limiting reactant.

<h3>3. <u>Answer</u>;</h3>

= 37.71 g

<h3><u>Explanation;</u></h3>

The balanced equation is:

2 Na + Cl₂ → 2 NaCl.

From the balanced equation:

2.0 moles of Na reacts with 1.0 mole of chlorine gas to give 2.0 moles of NaCl.

  • The number of moles of Na = mass/atomic mass),

                                                    n = (17.25 g / 22.989 g/mole of Na)

                                                         = 0.75 moles

Thus;

  • Using cross multiplication and from the equation:

                         2.0 moles of Na → 2.0 moles of NaCl

                         0.75 moles  of Na → 0.75 moles of NaCl

Therefore;

  • The mass of the produced NaCl (m = n x molar mass)

                    m = (0.75 mole x 58.443 g/mole of NaCl)

                         = 43.85 g.

  • This is the yield if the percentage of yield is 100%, but the percentage yield of the reaction is 86%.

Hence;

The actual yield = (43.85 g) (86/100)

                            = 37.71 g.

<h3>4. <u>Answer;</u></h3>

Oxygen and hydrogen molecules will mix because of random movement of their molecules.

<h3><u>Explanation;</u></h3>
  • Both gasses will diffuse into the other jar until the molecules are well distributed into both containers.
  • Diffusion is the movement of molecules from a point of high concentration to areas of low concentration until equilibrium is attained.  
  • The mixture of gases will have a partial pressure that is equal to the sum of the partial pressures of the component gases.
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Answer:

50.76 mol H2O.

Explanation:

The photosynthesis follows the equation:

6CO2 + 6H2O ---> C6H12O6 + 6O2

This means that 6 mol of H2O are needed to obtain 1 mol of C6H12O6 (see the numbers that precedes every molecule to know how many mols are in game).

So we can say that:

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8 0
3 years ago
A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a tempera
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Answer:

The final volume is 3.07L

Explanation:

The general gas law will be used:

P1V1 /T1 = P2V2 /T2

V2 =P1 V1 T2 / P2 T1

Give the variables to the standard unit:

P1 = 345 torr = 345 /760 atm = 0.4539atm

T1 = -15°C = -15 + 273 = 258K

V1 = 3.48L

T2 = 36°C = 36+ 273 = 309K

P2 = 468 torr = 468 * 1/ 760 atm = 0.6158atm

V2 = ?

Equate the values into the gas equation, you have:

V2 = 0.4539 * 3.48 * 309 / 0.6158 * 258

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3 years ago
The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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Answer:

The entropy decreases.

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