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3 years ago

12

If you drop a rubber ball covered in cooking sheet and fry it on a frying pan on hot temperature will it catch fire or will the

ball melt?

1 answer:

Anni [7]3 years ago

5 0

It will catch fire, the ball won't melt

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Water is an example of a compound. Which of the following is the best description of all compounds?

AlexFokin [52]

Answer:

A compound has atoms of different elements chemically joined together They can't be separated without a chemical reaction.

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2 years ago

The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi

Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the <em>length in millimetres</em>

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert <em>millimetres to metres </em>

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

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3 years ago

Water is heated to boiling and water vapor (steam) is released. Chemical change or physical change?

Vlad [161]

Physical because it is still H2O

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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

5 The apparatus shown below was set up. Give explanations for the following observations.

lana [24]

Answer:

hope it helps brainliest pls

Explanation:

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2 years ago