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3 years ago

The elements selenium and oxygen are both poor electrical conductors in which section of the periodic table are they found

Chemistry

2 answers:

Nutka1998 [239]3 years ago

6 0

Answer:

This question appears incomplete

Explanation:

The periodic table is divided into 4 blocks; s-block, p-block, d-block and the f-block.

Elements found in the s-block are metals; alkali metals and alkali earth metals. Elements found in the p-block include metals and non-metals while those found in the d-block are metals (transition metals). The F-block elements are the radioactive elements.

Oxygen (O) and selenium (Se) are non-metals found in group 16 of the periodic table which is found in the p-block of the periodic table. The p-block is located on the right side of the periodic table. Generally, non-metals are poor conductors of electricity.

Strike441 [17]3 years ago

5 0

Answer:

No metals. Right side.

Explanation:

On the right. Non metallic elements

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Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne

belka [17]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

$w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf$

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

$m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg$

c) m = 15 ton

$m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes$

4 0

3 years ago

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31

Anastaziya [24]

Answer:

The volume of the gas will be 78.31 L at 1.7 °C.

Explanation:

We can find the temperature of the gas by the ideal gas law equation:

$PV = nRT$

Where:

n: is the number of moles

V: is the volume

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

From the initial we can find the number of moles:

$n = \frac{P_{1}V_{1}}{RT_{1}} = \frac{1 atm*62.65 L}{(0.082 L*atm/K*mol)*(0 + 273)K} = 2.80 moles$

Now, we can find the temperature with the final conditions:

$T_{2} = \frac{P_{2}V_{2}}{nR} = \frac{612.0 mmHg*\frac{1 atm}{760 mmHg}*78.31 L}{2.80 moles*0.082 L*atm/(K*mol)} = 274.7 K$

The temperature in Celsius is:

$T_{2} = 274.7 - 273 = 1.7 ^{\circ} C$

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.

I hope it helps you!

8 0

3 years ago

The gasses in a hair spray can are at temperature 300k and a pressure of 30 atm, it

Sergeeva-Olga [200]

Answer:

900 K

Explanation:

Recall the ideal gas law:

$\displaystyle PV = nRT$

Because only pressure and temperature is changing, we can rearrange the equation as follows:
$\displaystyle \frac{P}{T} = \frac{nR}{V}$

The right-hand side stays constant. Therefore:

$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.

Substitute and solve for <em>T</em>₂:

$\displaystyle \begin{aligned} \frac{(30\text{ atm})}{(300\text{ K})} & = \frac{(90\text{ atm})}{T_2} \\ \\ T_2 & = 900\text{ K}\end{aligned}$

Hence, the temperature must be reach 900 K.

7 0

2 years ago

Each warning label must contain this element and its purpose is to provide a visual alert to users to the chemical hazards to wh

jekas [21]

The correct answer of the given question above would be a PICTOGRAM. OSHA’s required pictograms must be in the shape of a square set at a point and include a black hazard symbol on a white background with a red frame sufficiently wide enough to be clearly visible.

8 0

2 years ago

Read 2 more answers

Solve

stiv31 [10]

The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the following Charle's law equation:

V1/T1 = V2/T2

Where;

T1 = initial temperature
T2 = final temperature
V1 = initial volume
V2 = final volume

V1 = 9L
V2 = ?
T1 = 325K
T2 = 450K

9/325 = V2/450

325V2 = 4050

V2 = 4050/325

V2 = 12.46L

Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

Learn more about volume at: brainly.com/question/2817451

7 0

2 years ago