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3 years ago

How many electrons does a halogen need to comeplete an pctet?

1 answer:

7 0

Halogen have 7 valance electrons, for a halogen to be stable it need 8 valance electrons to be stable. So It needs 1 more valance electrons.

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If a solid has a density of 4.0 g/cm^3, what volume of the solid has a mass<br> of 24 g? Show work.

Mrrafil [7]

Answer:

<h2>Volume = 6 cm³</h2>

Explanation:

Density of a substance is given by

$Density = \frac{mass}{volume}$

From the question

Density = 4.0 g/cm³

mass = 24g

Substitute the values into the above formula and solve for the volume

That's

$4 = \frac{24}{v} \\4v = 24$

Divide both sides by 4

v = 6

We have the final answer as

<h3>Volume = 6 cm³</h3>

Hope this helps you

8 0

3 years ago

How does the smell of food being cooked spread so fast ???

sladkih [1.3K]

Answer:

When we cook food in the kitchen, that's the region of higher concentration of the smell. By diffusion, the smell spreads to the whole room and thereby whole house, so anyone standing at a distance, can smell it.

5 0

3 years ago

Read 2 more answers

An oil tanker spills a large amount of oil near the ocean shoreline. Which application of chemistry would best solve this proble

DochEvi [55]

I have heard they can use hair is stocking or nets to absorb the oil out of the water

3 0

3 years ago

On a coordinate plane, which single transformation could map point A (3,-4) onto point A' (1.5,-2)? *

Ksenya-84 [330]

Answer: A dilation with rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

Explanation:

A dilation is a non-rigid transformation that creates an image that is the same shape as the original but has a different size.

It uses a scale factor k such that

$(x,y)\to(kx,ky)$

(x,y)= coordinates of original figure

(kx,ky) = corresponding coordinate in the image.

To transform: A (3,-4) onto point A' (1.5,-2).

Using scale factor k= $\dfrac12$ , we have

$A(3,-4)\to A'(\dfrac12\times3, \dfrac12\times-4)=A'(1.5,-2)$

Required rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

3 0

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

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