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3 years ago

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Which statement about scientific explanations is true?

2 answers:

nadezda [96]3 years ago

8 0

Answer:

The correct option is A

Explanation:

Scientific explanations are made as a result of carefully researched and experimented procedures in order to arrive at a logical and probable conclusion on a subject matter. However, these scientific explanations are usually made based on facts as at the time it is been made. These facts could however be incomplete making them subject to change when more facts are produced.

For example, the Dalton's atomic theory became modified after new evidences emerged that atom is not the smallest unit of matter (with the discovery of electrons, protons and neutrons), atoms of the same elements are some times not alike in all aspect (the discovery of isotopes) among other modifications made to his theory.

This example among others indicate that scientific explanations are open to change/modification.

VashaNatasha [74]3 years ago

7 0

A. Scientific explanations are open to change.

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<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

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