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3 years ago

7

Sailin' Steve likes to test the buoyancy of different objects. He threw a spherical object into a fluid that has a greater densi

ty than the object does. He also threw a square object into the fluid that has a greater density than the fluid.

1 answer:

Hunter-Best [27]3 years ago

7 0

The spherical object will float and the square object will sink to the bottom of the fluid as it has greater density. :)

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When work is done on an object, the amount of ________ it has changes.

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When work is done on an object, the amount of energy it has changes.

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3 years ago

Swimming pool structures and structural _____ [680.26(B)(1) and (B)(2)] shall not be used as a grounding electrode.

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A grounding electrode is any object that directly links to the earth. They are most times used to divert electricity from the elements.

Swimming pool structures and structural <u>reinforcing steel. 250.52(B)(3)</u><u>,</u> [680.26(B)(1), and (B)(2)] shall not be used as a grounding electrode.

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2 years ago

This fine-grained dark colored igneous rock is most likely to

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3 years ago

35. In the collision theory, a collision that leads to the formation of products is called an

FinnZ [79.3K]

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3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago