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Shtirlitz [24]
3 years ago
13

An X-ray technician will place a lead cloth over parts of your body when taking an X-ray to protect your cells from being damage

d. Why do X-rays have the potential to harm your body but radio waves do not?
Chemistry
2 answers:
iren [92.7K]3 years ago
4 0
Radio waves have a lower frequency than x rays.
vichka [17]3 years ago
4 0

<u>Answer: </u>X-rays have shorter wavelength and high energy, therefore they affect more than radio waves.

<u>Explanation:</u>

Radio waves and X-rays are a type of electromagnetic waves which have both electrical and magnetic components associated with them.

X-rays have shorter wavelength than radio waves.

Relationship between the energy of the waves and the wavelength is given by Planck's equation, which is:

E=\frac{hc}{\lambda}

Where,

E = energy of the wave

h = Planck's constant

c = speed of light

\lambda = wavelength of the wave

From the above relation, energy is inversely related to the wavelength. As, the wavelength decreases, energy of the wave increases and vice-versa.

As, energy of the X-rays is more, they have high penetration power and hence, will harm the body more.

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What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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