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2 years ago

What is one similarity between metals and metalloids?

1 answer:

4 0

The correct answer that explains similarities between metal and Metalloids as regards the question is They both conduct electricity

Metalloids can be regarded as elements that are similar to metals, this is because they posses valence orbitals which is described as highly delocalized over macroscopic volumes.

As a result of this they can serve as electrical conductors.

metalloids posses small energy gap which is located between the valence band as well as the conduction band, as a result of this they are considered as intrinsic semiconductors when compare to pure conductors like metal.

Example of metal is Calcium, sodium and that of Metalloids are silicon and germanium

Therefore, metal and Metalloids are similar because of their conductivity of electricity

Learn more at: brainly.com/question/21036799?referrer=searchResults

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Macon is 50 miles from Atlanta. If the time of the trip increases, what happens to the speed? Question 8 options: A:It stays the

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The answer is B - it decreases.

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3 years ago

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Aqueous humor forms during capillary filtration in the __________?

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Answer:

Ciliary body.

Explanation:

Ciliary body: It is the known for the part of the eye that includes the ciliary muscle, which helps in the control the ciliary epithelium and lens shape, which are helping in the production of aqueous humor.

Through active secretion mechanism helping in to produce eighty percent of aqueous humor, and through the plasma ultra-filtration mechanism twenty percent of aqueous humor is produced.

Ciliary body is the part of the layer which helps to deliver the nutrients, and oxygen to the eye tissues, and this layer is known as uvea.

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3 years ago

When an alkane reacts with an element from group 7a, the reaction is referred to as?

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Alkanes are saturated hydrocarbon, that is they contain hydrogen and carbon without a double or triple bond between the carbon atoms, e.g. ethane, propane. Group 7a in the periodic table are called halogens e.g chlorine, bromine. Alkanes react with halogens in a reaction called substitution, where halogens replace hydrogen atoms in alkanes.

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3 years ago

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A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

aivan3 [116]

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = $n_1=9.79 mol$

Initial volume of freon gas in ballon = $V_1=?$

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = $n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol$

Final volume of freon gas in ballon = $V_2=21.8 L$

Using Avogadro's law:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ( at constant pressure and temperature)

$V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L$

16.06 L was the initial volume of the balloon.

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3 years ago

Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%

nlexa [21]

Answer: The empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

Explanation:

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Magnesium = $\frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles$

Moles of Silicon = $\frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.752 moles.

For Magnesium = $\frac{1.17}{0.752}=1.5$

For Silicon = $\frac{0.752}{0.752}=1$

For Hydrogen = $\frac{1.16}{0.752}=1.5$

For Oxygen = $\frac{3.07}{0.485}=4.08\approx 4$

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

Step 3: Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is $Mg_3Si_2H_3O_8$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

$n=\frac{520.8g/mol}{259g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}$

Hence, the empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

5 0

3 years ago