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Morgarella [4.7K]
1 year ago
10

What is the general name of the elements whose properties are intermediate between those of metals and non-metals ?

Chemistry
1 answer:
siniylev [52]1 year ago
6 0

The general name of the elements whose properties are intermediate between those of metals and non-metals is known as metalloids.

Metalloid, in chemistry, is defined as an imprecise term used to describe a chemical element that forms a simple substance having properties intermediate between those of a typical metal and a typical non-metal. Most commonly used examples of metalloids are arsenic, boron, germanium, silicon, antimony, tellurium and pollanium.

We can identify metalloids in the modern periodic table by locating a zig-zag line which separates metals from non-metals. The borderline elements which are metalloids are - boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po) - This elements exhibit properties similar to both metals and nonmetals and are called metalloids.

Learn more about metalloids from the link given below.

brainly.com/question/10139108

#SPJ4

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A 1.00 liter solution contains 0.31 M hydrocyanic acid and 0.40 M sodium cyanide. If 0.100 moles of barium hydroxide are added t
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Explanation:

acid --> HCN

base --> KCN

now

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base will react with an acid

so

OH- + HCN ---> CN- + H20

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HCN is used up , so number of moles of HCN will decrease

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A  false

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3 years ago
Why did Cyrus Field want to build a transatlantic telegraph?
Paul [167]

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5 0
2 years ago
This question involves two calculations. The answer to the first part will be
ozzi

Answer :

(a) The mass of Al_2O_3 produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of Al = 85.1 g

Molar mass of Al = 27 g/mol

First we have to calculate the moles of Al

\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol

Now we have to calculate the moles of Al_2O_3

The balanced chemical equation is:

4Al+3O_2\rightarrow 2Al_2O_3

From the reaction, we conclude that

As, 4 moles of Al react to give 2 moles of Al_2O_3

So, 3.15 moles of Al react to give \frac{2}{4}\times 3.15=1.58 mole of Al_2O_3

Now we have to calculate the mass of Al_2O_3

\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3

Molar mass of Al_2O_3 = 102 g/mole

\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g

Therefore, the mass of Al_2O_3 produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%

Therefore, the percent yield of the reaction is, 72.5 %

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What do the repetitive patterns in a mineral form
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