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3 years ago

Which electrons are the valence electrons of the atom? Helpppppppp

Chemistry

1 answer:

Arlecino [84]3 years ago

4 0

Explanation:

electrons in the highest energy orbitals which are able to contribute to Bonding are valence electrons,....... please mark me brainiest

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The center on your school’s basketball team is 6 ft 10 in tall. How tall is the player in millimeters (mm)?

svp [43]

The answer is 6 ft 10 inches in millimeters (mm) is 0.833 ft.

Given,

The center of the school's basketball team is 6 ft 10 inches tall.

We have to convert the height of the player from feet and inches to feet.

Using the conversion factor,

1 ft = 12 inches

or, 12inches/ 1 ft

Converting 6ft 10 inches to ft, we get;

10 inches × 1 ft/ 12inches

= 0.833 ft

Therefore 6 ft 10 inches in millimeters (mm) is 0.833 ft.

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

To learn more about Millimeter and Unit conversions, visit: brainly.com/question/26371870

#SPJ4

5 0

1 year 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

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

3 years ago

Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure

vekshin1

Check the solution in the attachment.

7 0

3 years ago

The liquid form of matter is usually more dense than its: solid form gas form both solid and gas forms

Julli [10]

<span>The liquid form of matter is usually more dense than its gas form. This is because liquid molecules are closer together compared to gas molecules. An exception, however, is water. Water's solid form or ice is less dense than its liquid form because of the orientation of hydrogen bonds that lowers its density.</span>

5 0

3 years ago

Which of the following pictures correctly show the structure of the H+ ion? Click on the picture to see each choice.

lutik1710 [3]

Answer: It's only one proton in the middle of the red circle there's nothing outside it.

Explanation: I'm not sure sorry :|

hope this helps

7 0

2 years ago

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