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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:

115.625^{\circ}\text{F}

Explanation:

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As heat gain and loss in the system is equal we have

m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}

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How many atoms of h are in 1 mol of (nh42so4?
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Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8
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Answer :  The average rate of consumption of H^+ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

As we are given:

\frac{d[Br^-]}{dt}=1.81M/s

Now we have to determine the average rate of consumption of H^+ during the same time interval.

As,

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

or,

-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times (1.81M/s)

\frac{d[H^+]}{dt}=2.17M/s

Thus, the average rate of consumption of H^+ during the same time interval is, 2.17 M/s

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