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3 years ago

Name 3 substances that are solids at room temperature. Do the same for liquids and gases.

Chemistry

1 answer:

arlik [135]3 years ago

8 0

Answer and Explanation:

Solids:

We can name some chemical elements that in nature are solids at room temperature:

Aluminium (Al): it is a dultile metal with a color between silver and white.

Sulfur (S₈): it is a pale yellow solid.

Zinc (Zn): it is a metallic brittle solid.

Liquids:

Some substances that are liquids at room temperature are:

Water (H₂O): it is a subtance composed by 2 atoms of hydrogen and 1 atom of oxygen. It is a clear, limpid, odorless liquid.

Bromine (Br₂): it is a homogenoeus halogen which is liquid at room temperature. It has a color beteween red and brown.

Mercury (Hg): this heavy metal is liquid at room temperature.

Gases:

Hydrogen (H₂): it is a colorless, odorless and hightly inflammable gas.

Oxygen (O₂): a diatomic gas, without odor neither color, with an important role in biological proceesses.

Nitrogen (N₂): other diatomic gas, which is inert (it does not react easily).

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Note that 5° of longitude or latitude equals about 563 km. Use this information and a ruler to create a scale to use with this m

Ann [662]

Answer:

A suitable scale, say 1 cm: 100 km can be used.

Explanation:

Thinking process:

The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.

For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:

1: 500

This is appreciable for the paper size.

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4 years ago

A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4

morpeh [17]

Answer: The molecular formula for the given organic compound is $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=21.33g$

Mass of $H_2O=4.366g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

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 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is $C_2H_2O_4$

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inysia [295]

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umka21 [38]

The answer is -2512.4 kJ

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3 years ago

What implications need to be considered when creating environmental policies?

Igoryamba

Answer:

How will it affect the environment? Is its benefit worth the cost?

Explanation:

when creating environmental policies you should always look into if it is worth it or not and how it will affect the environment

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3 years ago

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