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Ksju [112]
3 years ago
10

What is the best example of electromagnetic energy in everyday life

Chemistry
2 answers:
Natasha2012 [34]3 years ago
8 0

Answer:best

Explanation:

Best

sertanlavr [38]3 years ago
6 0
Electromagnetic Radiation. Do you listen to the radio, watch TV, or use a microwave oven? All these devices make use of electromagnetic waves. Radio waves, microwaves, visible light, and x rays are all examples of electromagnetic waves that differ from each other in wavelength.
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Ionic crystals are excellent insulators and can hold a large amount of heat before melting or boiling.
sweet-ann [11.9K]
It is a true fact that ionic crystals are excellent insulators and can hold a large amount of heat before melting or boiling. The correct option among the two options that are given in the question is the first option. Salt is a great example of ionic crystals and we know that it takes a huge amount of time to melt or boil.
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How many valence electrons does an atom of al possess?
Irina18 [472]

Answer:

3

Explanation:

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3 years ago
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
2 years ago
Please, I need at least accurate answers for my Sound Knowledge Check for my Science class.
Alla [95]

Answer:

1.A

2.A

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3 years ago
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Scientist can use trees to look at climates of the past. How? What information can they gather and how do they gather it? Explai
andrezito [222]

Explanation:

Scientist use trees a whole lot to look at climate of the past by examining tree rings.

These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.

Tree rings can be used to decipher the age of a tree.

  • These three rings can be used to interpret climatic patterns.
  • During a wet climate, the tree rings are more robust and bigger.
  • In a dry climate, the rings are thinner.
  • These alternating patterns can be used to decipher the climatic signatures in a tree.
  • Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.

learn more:

Climate change brainly.com/question/7824762

#learnwithBrainly

6 0
2 years ago
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