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Katena32 [7]
3 years ago
9

Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u

nit of measurement?
Jill
Bob
Kim
Steve
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

The answer is Bob

Explanation:

The science exist to kinds of measurements base units and derive units of measurement.

Base units are the ones that measure: length, mass, time, electric current, temperature, luminous intensity and amount of substance.

Derivative units of measurement are obtain when we combine the base units.

Know from your question

Bob measures length is a base unit

Jill measures density is a derivative unit, is the combination of mass and volumen

Kim measures mass is a base unit

Steve measures brightness is a base unit.

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Explanation:

<h3>The given statement is related to the collision theory - </h3>

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Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3
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Solution : Given,

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First we have to calculate the moles of CH_4 and O_2.

\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 0.026 moles of O_2 react with \frac{0.026}{2}=0.013 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2

From the reaction, we conclude that

As, 2 mole of O_2 react to give 1 mole of CO_2

So, 0.026 moles of O_2 react to give \frac{0.026}{2}=0.013 moles of CO_2

Now we have to calculate the mass of CO_2

\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2

\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g

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Now we have to calculate the percent yield of CO_2

\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100

\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%

Therefore, the percent yield of CO_2 is, 68.4 %

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