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Anna007 [38]
3 years ago
13

A student reports the density of an article to be 1.92 g/mL. The accepted value of density is 1.89 g/mL. Calculate the student's

percent error.
Chemistry
1 answer:
ipn [44]3 years ago
6 0

Answer:

1.59%.

Explanation:

The following data were obtained from the question:

Measured density = 1.92 g/mL

Accepted density = 1.89 g/mL

Percentage error =...?

The percentage error error of the student can be obtained as follow:

Percentage error = |Measured density – Accepted density |/Accepted density × 100

Percentage error = |1.92 – 1.89|/1.89 × 100

Percentage error = 0.03/1.89 × 100

Percentage error = 1.59%

Therefore, the student percentage error is 1.59%

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Which actions are part of a system of methodical tests and refinements specific to technological design? 1 documenting, tinkerin
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Answer:

1. documenting, tinkering, testing

Explanation:

Technological design is defined as the process of study, design and development of new technologies.

There are some action in the methodical tests and refinements specific to technological design include documenting, tinkering, testing.

<u>Documenting </u><u>includes collecting all the information about the design and develop the product, </u><u>tinkering</u><u> involves repairing or adjust the issues found in the development, and </u><u>testing </u><u>helps to evaluate if the product is ready to work as it is supposed to.</u>

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3 years ago
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Answer:

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
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When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

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No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

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Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

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Definitely C, ability to create delicate flavors. :)
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