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3 years ago

5

In a particle accelerator, the accelerated particle primarily gains what?

1 answer:

ki77a [65]3 years ago

6 0

In a particle accelerator, the accelerated particle primarily gains kinetic energy. A particle accelerator is a device that accelerates particles which means it would add more energy to the particles increasing the kinetic energy of these particles. This is commonly used in medical and any scientific research.

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Which type of environmental scientist is likely to study how different species of birds interact?

Reptile [31]

Answer: D. Ecologist

Explanation: an ecologist is someone who studies the interactions of different organisms. An ecologist would be a type of scientist who would most likely study the interactions of birds

3 0

3 years ago

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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

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

<u>Explanation:</u>

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 %

5 0

3 years ago

Select the correct answer.

solong [7]

I think it’s A but I’m not sure, if it’s wrong I’m sorry

7 0

2 years ago

Is boiling eggs chemical or physical practice?

Elden [556K]

Answer:

chemical

Explanation:

because heat is being taken to the egg

3 0

3 years ago

What is the volume of 88.2 g of silver metal (density = 10.50 g/cm^3)

babymother [125]

Answer:

The answer is

<h2>8.4 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass = 88.2 g

density = 10.5 g/cm³

The volume is

$volume = \frac{88.2}{10.5}$

We have the final answer as

<h3>8.4 mL</h3>

Hope this helps you

8 0

3 years ago

Read 2 more answers