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4 years ago

What is required to shield alpha particles?

1 answer:

5 0

Shielding alpha and beta particles is part of radiation safety, which is followed very seriously in nuclear power plants, hospitals, laboratories where particle accelerators are used, and in many more situations. Alpha particles are much easier to block than beta particles. Alpha particles are positively charged helium nuclei, and beta particles are negatively charged electrons. Finding materials to block alpha and beta particles requires some serious consideration because of how dangerous radiation exposure is. Density is very important when is comes to blocking these particles. The cool thing about alpha particles, though, is how easily they can be blocked. You would think that blocking a particle would take some seriously thick metal or something, but you could actually use paper, or even plastic to block these particles!

Hopefully this was helpful to you - feel free to comment if you have any other questions!

You might be interested in

A teacher makes the following statement.

DedPeter [7]

I believe the correct answer from the choices listed above is option A. The topic that the teacher is talking about would be distillation of a mixture. Gasoline is processed by distillation. Hope this answers the question. Have a nice day.

5 0

3 years ago

Read 2 more answers

When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Temka [501]

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

4 0

3 years ago

How does increasing the slope of a ramp affect how far in marble travels?

zvonat [6]

The higher the slope of a ramp the more velocity and speed the marvel will gain, which means more distance traveled

7 0

3 years ago

How many molecules are contained in 103.4g of sulfuric acid?

Ierofanga [76]

<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Compounds</u>

Writing Compounds
Acids/Bases

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 103.4 \ g \ H_2S(\frac{1 \ mol \ H_2S}{34.09 \ g \ H_2S})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2S}{1 \ mol \ H_2S})$
Multiply: $\displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S

4 0

3 years ago

At STP, which 2.0-gram sample of matter uniformly fills a 340-milliliter closed container?

Xelga [282]

Answer;

D, Xe (g)

Solution and explanation;

If 2g has a volume of 340ml.

Density is 1000/340*2 = 5.88g/litre.

-This rules out the two solids, choices 2) &3)

If 1 litre has mass 5.88g,

then 22.4 liters (volume at STP) has mass 5.88*22.4 = 131.8g/mol

molar mass Br2 = 80*2 = 160g/mol NO

molar mass Xe = 131.3g/mol = YES.

Answer is Xe

6 0

3 years ago