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4 years ago

What elemental classification does oxygen belong to?

2 answers:

6 0

The elemental classification for oxygen would be non-metal. It is located in period number 2 and group number 16. It has an atomic number of 8. At room temperature, it exists as a gas. Hope this answers the question. Have a nice day.

Komok [63]4 years ago

3 0

Answer: Oxygen is a non-metal.

Explanation:

Elements are distributed into 7 period and 18 Groups in a periodic table.

Elements are classified into three categories:

Metals: They are called as metals because they loose electrons to attain stability. These elements belong from Group 1 to group 13 of the periodic table. They are hard and are mainly solid.
Non-metals: They are called as non-metals because they gain electrons to attain stability. These elements belong from Group 15 to group 17 of the periodic table. They are soft and are mainly liquid or gases.
Metalloids: They are called as metalloids because they show properties of both metals and non-metals. These elements belong from Group 14 of the periodic table.

Oxygen belongs to Group 16 and period 2 of the periodic table.

Hence, oxygen is a non-metal.

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3 years ago

i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t

Ronch [10]

Assuming that the number of mols are constant for both conditions:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Now you plug in the given values. V_1 is the unknown.
$\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}
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Separate V_1
$V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }$
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3 years ago

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3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, $q=mc\Delta T$ .

where, q is the heat energy, m is mass, c is specific heat and $\Delta T$ is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is $\frac{1.035cal}{g.^0C}$ .

$\Delta T$ for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

$\Delta T$ for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.

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4 years ago

Read 2 more answers

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The cell membrane which is more fat permeable will be present at the end of the process.

Solvent extraction is a very important method if separating mixtures. The principle of solvent extraction is based on the idea of like dissolves like. A substance dissolves in the component of the system in which it is most soluble.

If a cell extract is dissolved in a nonpolar solvent, the cell membrane which is mostly permeable to nonpolar molecules will dissolve most in the solvent.

Learn more:brainly.com/question/14396802

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3 years ago