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ahrayia [7]
4 years ago
6

_________ 1. an allotrope of carbon that is soft and can be used as a lubricant is ______.

Chemistry
1 answer:
dangina [55]4 years ago
8 0
<span>1. An allotrope of carbon that is soft and can be used as a lubricant is </span><span>d. graphite.
</span>Graphite is the most stable allotrope of carbon under standard conditions (pressure and temperature). In graphite carbon atoms have sp2 hybridization, because of that graphite has ability to conduct electricity and heat.

2. The process by which a solid changes directly to a gas without first becoming a liquid is called <span>c. sublimation.
</span>Sublimation is an endothermic process. For example dry ice (carbon(IV) oxide in solid state) is used because of sublimation in nightclubs, fog machines<span>, at </span>theaters<span>, </span>haunted house attractions.

3. In the gaseous state d. halogens form diatomic covalent molecules.
Halogen diatomic molecules (17 group<span> in the </span>periodic table, halogen means "salt producing) in the gaseous state in room temperature are fluorine (F₂) and chlorine (Cl₂). Both molecules have single covalent bond and they are very reactive.

4. When hydrogen reacts with the active metals, it <span>b. gains one electron.
Active metals are I group of Periodic table of elements (alkaline metals). Alkaline metals are </span><span>lithium, sodium, rubidium, potassium, cesium. They have low ionization energy and lose one electron very easy. Compounds with hydrogen with this metals are called hydrides.

5. D</span>ifferent forms of the same element that have different properties because of different atom arrangements are called <span>a. allotropes.
Carbon has many allotropes, but two most important are graphite and diamomd. Graphite has sp2 and diamond has sp3 hybridization of carbon atoms, because of that graphite conduct electricity and diamond not.

6. A</span>t room temperature, most metals are <span>b. solids.
Most metals have strong metallic bond, because strong </span><span>electrostatic attractive force between </span>valence electrons (metals usually have low ionization energy and lose electrons easy) <span>and positively charged </span>metal<span> ions.
</span>They also form <span>crystal lattice and because of that has high melting points.
</span><span>
7. A</span>n allotrope of carbon that is hard and is often used in jewelry is b. diamond.
In diamond carbon atoms<span> are arranged in the </span>face centered cubic<span> crystal structure called a </span>diamond lattice.
Diamond has very strong covalent bonds between carbon atoms and because of that <span>it has the highest </span>hardness<span> and </span>thermal conductivity<span> of any bulk substance. 

8. M</span>etals can be used as wire because they are a. ductile.
Ductility<span> is a measure of a material's (in this example metal) ability to undergo deformation before rupture. For example g</span><span>old is very ductile and can be drawn into a monoatomic wire.</span>
Malleability<span> is a material's ability to deform under </span>compressive<span> stress.
</span>
9. The d. transition elements often occur in nature as uncombined elements.
Uncombined elements are in elemental form and do not attache with other elements. Several transition elements (metals) like gold, silver, rhodium, palladium and platinum are unreactive and <span>are resistant to oxidation and corrosion.</span>
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Draw the Lewis structure (including all lone pair electrons) for one isomer of the molecular formula C3H8O.
Andreyy89

Answer:

Here's how I would do it.

Explanation:

A Lewis structure shows the valence electrons surrounding the atoms.

1. Identify the central atoms

The central atoms are the least electronegative atoms (C).

I chose as my central atoms a chain of four C atoms: C-C-C-C

I stuck the most electronegative atom (O) at the end of the chain:

C-C-C-C-O

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3 years ago
How many molecules are there in 2.30 grams of NH3 ?
Evgesh-ka [11]
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3 years ago
Electrochemistry - Equilibrium
Ipatiy [6.2K]

Answer:

Explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK  , F is faraday constant T is 273 + 25 = 298 K

E⁰cell  =  1.46 - 1.21 = .25 V

n = 2

Putting the values

.25 = (8.314 x 298  lnK) / (2 x 96485 )

lnK = 19.47

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2 )

Change in free energy Δ G

Δ G ⁰ = nE⁰ F

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E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

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3 )

E⁰cell = (RT / nF ) lnK

n = 2

1.78 = 8.314 x 298  lnK / 2 x 96485

lnK = 138.638

K = 1.62 x 10⁶⁰

8 0
3 years ago
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