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ozzi
4 years ago
10

If an element is lustrous, brittle, and a semi-conductor, how would you classify it?

Chemistry
1 answer:
just olya [345]4 years ago
7 0

Metalloid

Explanation:

If an element is lustrous, brittle and a semi-conductor, it is best classified as a metalloid.

Metalloids shares attributes of metals and non-metals.

  • They are often described as semi-metals as they do not share the full properties that makes a metal a metal.
  • Metalloids are lustrous but not malleable like metals.
  • They do not conduct electricity but they do so on certain conditions.
  • Examples are silicon, germanium, boron, arsenic e.t.c
  • They are usually found in the middle of the periodic table.
  • They are not readily alloyed with metals.

Learn more:

Metalloid brainly.com/question/3023499

#learnwithBrainly

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Which type of particles have no definate shape, but they do have a definite volume?
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Liquid, liquids have a definite volume but can take the shape of whatever container holds them.
6 0
3 years ago
If 10 grams of NaHCO3 react with an excess of HCl, what would be the theoretical mass of NaCl produced?
melamori03 [73]

Answer:

Explanation:Reaction NaHCO3 + HCl ⇒ NaCl + H2O + CO2

Molar mass M = (22.99+1.008+ 12.01+3·16 ) g/mol. Calculate

Amount of substance n =m/M, n(NaCl) is equal.

M(NaCl) = 58.44 g/mol and mass m= n·M

6 0
3 years ago
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)? Show all calculations leading to an answer. Li
Vaselesa [24]

Balanced chemical reaction happening here is:

3Mg(s) + N₂(g) → Mg₃N₂(s)        


 <u>moles of product formed from each reactant:</u>


2.0 mol of N2 (g) x <u> 1 mol Mg₃N₂      </u>  = <u>2 mol Mg₃N₂</u>

                                    1 mol N2

and


8.0 mol of Mg(s) x <u> 1 mol Mg₃N₂      </u>   = 2.67 mol Mg₃N₂

                                 3 mol Mg


Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂

N2 is the limiting reactant here and Mg is excess reactant.


Hence mole of product formed here is 2 mol Mg₃N₂    


molar mass of Mg₃N₂    

= 3 Mg + 2 N

= 101g/mol  


mass of product(Mg₃N₂) formed  

= moles x Molar mass

= 2 x 101

= 202g Mg₃N₂


<u>202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).</u>


<u>   </u>   The following are indicators of chemical changes:

Change in Temperature    

Change in Color

Formation of a Precipitate



8 0
3 years ago
Two substances in a mixture differ in density and particle size. theses properties can be used to
kykrilka [37]

Answer:

option 1. Two substances in a mixture differ in density and particle size. These properties can be used to separate the substances. These properties can be manipulated in order to have a better separation between the two substances.

Explanation:

8 0
3 years ago
Read 2 more answers
Compounds A and B are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 g of sulf
sukhopar [10]

Answer:

1.5

Explanation:

Given that :

Compound A and B are formed from Sulfur + Oxygen.

Compound A :

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

Comparing the ratios :

Compound A:

S : O = 6.00 : 5.99

S/0 = 6.0g S / 5.99g O

Compound B :

S : O = 8.60 : 12.88

S / O = 8.60g S / 12.88g O

Mass Ratio of A / mass Ratio of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

= 77.28 / 51.514

= 1.50017

= 1.5

4 0
4 years ago
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