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Alla [95]
3 years ago
12

When making observations during an experiment, which of the following must be included? a. Descriptions of the appearance of any

chemicals used. b. Explanations for why certain steps were carried out analysis of results. c. Descriptions of any changes in the system following an action. d. Descriptions of any non-changes in the system following an action a list of possible errors.
Chemistry
1 answer:
sweet [91]3 years ago
8 0

Answer:When making observations during an experiment, Descriptions of any changes in the system following an action and Descriptions of any non-changes in the system following an action a list of possible errors must be included. The correct option is C and D.

Explanation: During a scientific experiment procedures, observation is usually carried out to take into consideration an active information from the natural source. Data can be collected by the observer:

- quantitatively ( by counting and measuring data when numerical values are attached.

- qualitatively ( when the presence or absence of a property is noted). Hence, option C and D are correct.

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Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen
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Answer:

5.9 kg  

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ:                                     55.84

            Fe₂O₃ + 3CO  ⟶  2Fe  +  3CO₂

m/kg:                                    7.0

(a) Moles of Fe

\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}

(b) Moles of CO

\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ:                32.00

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n/mol:                             261

(a) Moles of O₂

\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}

(b) Mass of O₂

\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol  O}_{2}} = \text{4180 g O}_{2}

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

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