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kifflom
3 days ago
5

Name one manufactured device or natural phenomenon that emits electromagnetic radiation in each of the following wavelengths: ra

dio, microwave, infrared, visible light, ultraviolet, X-ray, and gamma ray.
Chemistry
1 answer:
alisha [964]3 days ago
7 0
Radio - A radio station sends out radio waves that are received by a radio receiver.
<span>Microwaves - A microwave oven heats food using microwave radiation.</span>

<span>Infrared - Infrared light is utilized by TV remotes to send signals to a sensor on the TV, enabling functions like volume adjustment and channel selection.</span>

<span>Visible light - Comes from sunlight or light bulbs.</span>

<span>Ultraviolet - UV lamps are used for tanning and for verifying the authenticity of currency.</span>

<span>X-rays - Machines for chest X-rays and backscatter X-ray scanners for airport security utilize X-rays.
</span>
Gamma rays - <span>Gamma rays are utilized in medical instruments for eliminating cancer cells and for sterilizing medical supplies.</span>
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(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [854]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
16 days ago
A certain liquid has a density of 2.67 g/ cm3. what is the mass of 30.5 ml of this liquid? (
lorasvet [960]
Hello!

density = 2.67 g/cm³

volume = 30.5 mL

Thus:

Mass = density * volume

Mass = 2.67 * 30.5

Mass = 81.435 g 
4 0
7 days ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
eduard [944]

Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

5 0
8 days ago
At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C
VMariaS [1037]

Answer: 0.0007 moles of CO_2 are released when the temperature rises.

Explanation:

To determine the moles, we utilize the ideal gas law, as follows:

PV=nRT

where,

P = gas pressure = 1.01 bar

V = gas volume = 1L

R = gas constant = 0.08314\text{ L bar }mol^{-1}K^{-1}

  • Calculated moles at T = 20° C

The gas temperature = 20° C = (273 + 20)K = 293K

Substituting values into the equation gives:

1.01bar\times 1L=n_1\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 293K\\n_1=0.04146moles

  • Calculated moles at T = 25° C

The gas temperature = 25° C = (273 + 25)K = 298K

Substituting values into the equation gives:

1.01bar\times 1L=n_2\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 298K\\n_2=0.04076moles

  • Released moles = n_1-n_2=0.04146-0.04076=0.0007moles

Therefore, 0.0007 moles of CO_2 are released when the temperature increases from 20° C to 25° C.

5 0
13 days ago
A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is pr
VMariaS [1037]
A mixture is created by dissolving 1.43 mol of potassium chloride (KCl) in 889 g of water. The concentration of KCl works out to be 1.61 molal.
The amount of KCl is 1.43 mol
Water weighs 889 g
The molality can be calculated using the formula:
molality = moles of solute divided by kilograms of solvent
Since 1 kg equals 1000 g, 889 g is 0.889 kg.

Therefore, m = 1.43/0.889 = 1.61 molal.
7 0
4 days ago
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