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Alisiya [41]
3 years ago
8

A substance that forms a vapor is generally in what physical state at room temperature? A gas b liquid c liquid or solid d solid

Chemistry
1 answer:
Korvikt [17]3 years ago
6 0

B liquid hope this helps :)

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Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.
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When heating a sample to extreme heat over a flame, you will often use a crucible. Before starting the heating, you should locat
Ghella [55]

Answer:

The equipments you should have ready to start the crucible experiment includes: safety goggles, crucible with lid, crucible tong, ring support with clay triangle, Bunsen burner and heat resistant tile.

Explanation:

Crucible is an equipment in the laboratory which is suitable for heating a sample to extreme heat over a flame, Modern laboratory crucible are made up of graphite- based composite materials for achievement of higher performance. Because extreme heat is involved, you should locate the correct labware for the experiment, including the equipment to safely handle and support the crucible. These equipments includes:

--> Safety goggles: Because you will work with chemical it is advisable to use a safety goggles which protects the eyes from dangerous floating chemical aerosol.

--> crucible with lid: This is the main apparatus with the lid (cover) which is used to cover the crucible to prevent spilling of the boiling chemical.

--> Crucible tong: These are scissors like tools used to grasp hot crucible.

--> Ring support with clay triangle: the clay triangle is used to hold crucible when they are being heated. They usually sit on a ring stand.

--> Bunsen burner: Produces a single open gas flame which can be used for heating.

With the safety equipments listed above, you can carry out experiment using the crucible. These equipments helps minimise laboratory hazard that may occur should Incase it's not available.

6 0
3 years ago
Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%
nlexa [21]

<u>Answer:</u> The empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

<u>Explanation:</u>

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:

  • <u>Step 1:</u> 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

  • <u>Step 2:</u> 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

  • <u>Step 3:</u> 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
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